Foundations of Electromagnetic Theory 611 spring 19...Electrostatic Theory: 3 charges are not...

1Prof. Sergio B. MendesSpring 2018

Prologue of“Modern Problems in Classical Electrodynamics”

by Charles Brau

Foundations of Electromagnetic Theory

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• Topics to be covered in this prologue: electrostatics, magnetostatics, electrodynamics, electromagnetic waves, conservation laws, and Maxwell’s stress tensor.

• Almost all forces perceived in Nature (except for gravity) are electromagnetic forces.

• For most of the topics listed above, this is intended to be a review, not an introduction.

Prof. Sergio B. MendesSpring 2018

Foundations of Electromagnetic Theory

Electrostatic Theory:

3

charges are not moving, they are fixed in space.

charge density: 𝜌𝜌 𝒓𝒓, 𝑡𝑡 = 𝜌𝜌 𝒓𝒓


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Milestones in Electrostatics• 600 B.C.: Thales of Miletus, amber rubbed with fur attracts other

object, electron = amber in greek

• 1600 D.C.: Gilbert, “amberness” phenomena is also displayed by other materials

• 1733: du Fay, repulsion is also possible (in addition to attraction), two flavors of electric charge

• 1834: Faraday, electric charge comes in discrete amounts (is quantized in modern terminology)

• 1909: Millikan, 𝑒𝑒 ≅ −1.62 × 10−19 𝐶𝐶

• 1746: Watson, 1747: Franklin, electricity as a fluid that moves from one body to another, conservation of electric charge

• 1785: Coulomb, force between small electrically-charged objects


5

𝑭𝑭𝒒𝒒 𝒓𝒓 =𝑞𝑞 𝑞𝑞𝑞

4 𝜋𝜋 𝜖𝜖0𝒓𝒓 − 𝒓𝒓′

𝒓𝒓 − 𝒓𝒓𝑞 3

Coulomb Law

𝑞𝑞𝑞𝑞𝑞

𝒓𝒓𝒓𝒓𝑞

𝒓𝒓 − 𝒓𝒓𝑞𝑭𝑭

𝒪𝒪

(force on 𝑞𝑞 located at 𝒓𝒓due to 𝑞𝑞𝑞 located at 𝒓𝒓𝑞)

𝑭𝑭𝒒𝒒 𝒓𝒓

∝ 𝑞𝑞 𝑞𝑞𝑞

𝜖𝜖0 ≅ 8.854187817 × 10−12𝐶𝐶2

𝑁𝑁 𝑚𝑚2

(permittivity of free space)Prof. Sergio B. MendesSpring 2018

∝𝟏𝟏


6

𝑭𝑭𝒒𝒒 𝒓𝒓 =𝑞𝑞

4 𝜋𝜋 𝜖𝜖0�𝑖𝑖 = 𝑎𝑎𝑎𝑎𝑎𝑎

𝑞𝑞𝑖𝑖𝒓𝒓 − 𝒓𝒓𝑖𝑖𝒓𝒓 − 𝒓𝒓𝑖𝑖 3

𝑭𝑭𝒒𝒒 𝒓𝒓 =𝑞𝑞

4 𝜋𝜋 𝜖𝜖0�−∞

+∞

𝜌𝜌 𝒓𝒓𝑞 𝑑𝑑𝑑𝑑𝑞𝒓𝒓 − 𝒓𝒓′

𝒓𝒓 − 𝒓𝒓′ 3

𝑑𝑑𝑞𝑞 𝒓𝒓𝑞 = 𝜌𝜌 𝒓𝒓𝑞 𝑑𝑑𝑑𝑑𝑞

• Continuous distribution of charges 𝜌𝜌 𝒓𝒓𝑞 :

• Discrete charges 𝑞𝑞𝑖𝑖 located at 𝒓𝒓𝑖𝑖:

Net Force from Multiple ChargesPrinciple of Superposition: forces are added (vectorially)


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The force 𝑭𝑭𝒒𝒒 𝒓𝒓 on 𝑞𝑞 is linearly proportional to 𝑞𝑞:

The proportionality constant is called the electric field 𝑬𝑬 𝒓𝒓 :

𝑬𝑬 𝒓𝒓 ≡𝑭𝑭𝒒𝒒 𝒓𝒓𝒒𝒒


An Important Consequence:

𝑭𝑭𝒒𝒒 𝒓𝒓 ∝ 𝑞𝑞

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𝑬𝑬 𝒓𝒓 =1




4 𝜋𝜋 𝜖𝜖0�−∞

+∞

𝜌𝜌 𝒓𝒓𝑞 𝑑𝑑𝑑𝑑′𝒓𝒓 − 𝒓𝒓′

𝒓𝒓 − 𝒓𝒓′ 3

• Multiple discrete charges 𝑞𝑞𝑖𝑖 located at 𝒓𝒓𝑖𝑖:

• Continuous distribution of charges 𝜌𝜌 𝒓𝒓𝑞 :

• Single charge 𝑞𝑞𝑞 located at 𝒓𝒓𝑞:

𝑬𝑬 𝒓𝒓 =𝑞𝑞𝑞



Electric Field


𝒓𝒓

𝒪𝒪

𝒓𝒓

𝒪𝒪

𝒪𝒪

𝒓𝒓𝑞 𝑞𝑞𝑞 𝒓𝒓

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Flux of the Electric Field on a Arbitrary Surface

𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝑬𝑬 ≡�𝑆𝑆𝑬𝑬 𝒓𝒓 . 𝒅𝒅𝒅𝒅

𝑬𝑬 𝒓𝒓

𝒅𝒅𝒅𝒅𝑆𝑆

𝑬𝑬 𝒓𝒓 . 𝒅𝒅𝒅𝒅 = 𝑬𝑬 𝒓𝒓 cos 𝜃𝜃 𝒓𝒓 𝑑𝑑𝑆𝑆

𝜃𝜃 𝒓𝒓

𝒪𝒪

𝒓𝒓

vector vectordot

product

𝒅𝒅𝒅𝒅 ≡ �𝒏𝒏 𝒓𝒓 𝑑𝑑𝑆𝑆

𝑑𝑑𝑆𝑆

vector unit normal vector

differential area


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Divergence Theorem

�𝑆𝑆𝑸𝑸 𝒓𝒓 . 𝒅𝒅𝒅𝒅 = �

𝑉𝑉𝛻𝛻.𝑸𝑸 𝒓𝒓 𝑑𝑑𝑑𝑑

Consider a closed surface S that surrounds a volume V

also known as Gauss's theorem or

Ostrogradsky's theorem


S

V

S

V

SV

and a vector field 𝑸𝑸 𝒓𝒓 , if 𝑸𝑸 𝒓𝒓 is a continuously differentiable vector field:

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Illustration: Net Flux along x-direction


𝑸𝑸 𝒓𝒓 + ∆𝐹𝐹 �𝒆𝒆𝒙𝒙

= 𝑸𝑸 𝒓𝒓 + ∆𝐹𝐹 �𝒆𝒆𝒙𝒙 . + �𝒆𝒆𝒙𝒙 + 𝑸𝑸 𝒓𝒓 . − �𝒆𝒆𝒙𝒙 ∆𝑦𝑦 ∆𝑧𝑧

𝑸𝑸 𝒓𝒓

𝐹𝐹

𝑦𝑦

𝑧𝑧𝑸𝑸 𝒓𝒓 . 𝒅𝒅𝒅𝒅

= 𝑄𝑄𝑥𝑥 𝒓𝒓 + ∆𝐹𝐹 �𝒆𝒆𝒙𝒙 − 𝑄𝑄𝑥𝑥 𝒓𝒓 ∆𝑦𝑦 ∆𝑧𝑧

∆𝐹𝐹

=𝜕𝜕𝑄𝑄𝑥𝑥𝜕𝜕𝐹𝐹 ∆𝐹𝐹 ∆𝑦𝑦 ∆𝑧𝑧

= 𝑄𝑄𝑥𝑥 𝒓𝒓 +𝜕𝜕𝑄𝑄𝑥𝑥𝜕𝜕𝐹𝐹 ∆𝐹𝐹 − 𝑄𝑄𝑥𝑥 𝒓𝒓 ∆𝑦𝑦 ∆𝑧𝑧

∆𝑦𝑦

∆𝑧𝑧

+�𝒆𝒆𝒙𝒙

−�𝒆𝒆𝒙𝒙

𝒅𝒅𝒅𝒅 = ∆𝑦𝑦 ∆𝑧𝑧 ± �𝒆𝒆𝒙𝒙

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Net Flux along all directions


𝑸𝑸 𝒓𝒓

𝐹𝐹

𝑦𝑦

𝑧𝑧

𝑸𝑸 𝒓𝒓 . 𝒅𝒅𝒅𝒅

∆𝐹𝐹

=𝜕𝜕𝑄𝑄𝑥𝑥𝜕𝜕𝐹𝐹

+𝜕𝜕𝑄𝑄𝑦𝑦𝜕𝜕𝑦𝑦

+𝜕𝜕𝑄𝑄𝑧𝑧𝜕𝜕𝑧𝑧

∆𝐹𝐹 ∆𝑦𝑦 ∆𝑧𝑧

∆𝑦𝑦

∆𝑧𝑧

= 𝛻𝛻.𝑸𝑸 𝑑𝑑𝑑𝑑

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Electric Flux on a closed surface:

𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝑬𝑬, 𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒆𝒆𝒅𝒅 = �𝑆𝑆𝑬𝑬 𝒓𝒓 . 𝒅𝒅𝒅𝒅

𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝑬𝑬,𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒆𝒆𝒅𝒅 = �𝑆𝑆𝑬𝑬 𝒓𝒓 .𝒅𝒅𝒅𝒅 = �

𝑉𝑉𝛻𝛻.𝑬𝑬 𝒓𝒓 𝑑𝑑𝑑𝑑

Divergence Theorem


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Single charge 𝑞𝑞𝑞 located at 𝒓𝒓𝑞:

𝑬𝑬 𝒓𝒓 =𝑞𝑞𝑞



𝛻𝛻−1𝒓𝒓 − 𝒓𝒓𝑞

=𝒓𝒓 − 𝒓𝒓′


�𝑆𝑆𝑬𝑬 𝒓𝒓 .𝒅𝒅𝒅𝒅 = �

𝑉𝑉𝛻𝛻.𝑬𝑬 𝒓𝒓 𝑑𝑑𝑑𝑑 = �

𝑉𝑉

𝑞𝑞𝑞4 𝜋𝜋 𝜖𝜖0

𝛻𝛻. 𝛻𝛻−1𝒓𝒓 − 𝒓𝒓𝑞

𝑑𝑑𝑑𝑑

HW:

HW:

=𝑞𝑞𝑞

4 𝜋𝜋 𝜖𝜖0𝛻𝛻

−1𝒓𝒓 − 𝒓𝒓𝑞

𝛻𝛻. 𝛻𝛻𝜓𝜓 𝒓𝒓 = 𝛻𝛻2 𝜓𝜓 𝒓𝒓

= �𝑉𝑉


𝛻𝛻2−1𝒓𝒓 − 𝒓𝒓𝑞

𝑑𝑑𝑑𝑑


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𝛻𝛻2−1𝒓𝒓 − 𝒓𝒓𝑞

= 4 𝜋𝜋 𝛿𝛿3 𝒓𝒓 − 𝒓𝒓𝑞HW:

𝑞𝑞′

𝜖𝜖0

0

if q’ is inside V

if q’ is outside V

= �𝑉𝑉

𝑞𝑞′

4 𝜋𝜋 𝜖𝜖04 𝜋𝜋 𝛿𝛿3 𝒓𝒓 − 𝒓𝒓𝑞 𝑑𝑑𝑑𝑑


𝑉𝑉


𝛻𝛻2−1𝒓𝒓 − 𝒓𝒓𝑞

𝑑𝑑𝑑𝑑 =


{=

=𝑞𝑞′

𝜖𝜖0�

𝑉𝑉𝛿𝛿3 𝒓𝒓 − 𝒓𝒓𝑞 𝑑𝑑𝑑𝑑

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Multiple discrete charges 𝑞𝑞𝑖𝑖 located at 𝒓𝒓𝑖𝑖:




=1


𝑞𝑞𝑖𝑖 𝛻𝛻−1

𝒓𝒓 − 𝒓𝒓𝑖𝑖

= �𝑉𝑉

14 𝜋𝜋 𝜖𝜖0

�𝑖𝑖 = 𝑎𝑎𝑎𝑎𝑎𝑎

𝑞𝑞𝑖𝑖 𝛻𝛻2−1

𝒓𝒓 − 𝒓𝒓𝑖𝑖𝑑𝑑𝑑𝑑

= �𝑉𝑉

14 𝜋𝜋 𝜖𝜖0

�𝑖𝑖 = 𝑎𝑎𝑎𝑎𝑎𝑎

𝑞𝑞𝑖𝑖 4 𝜋𝜋 𝛿𝛿3 𝒓𝒓 − 𝒓𝒓𝑖𝑖 𝑑𝑑𝑑𝑑

= �𝑖𝑖= 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖

𝑞𝑞𝑖𝑖𝜖𝜖0

= �𝑖𝑖=𝑎𝑎𝑎𝑎𝑎𝑎


�𝑉𝑉𝛿𝛿3 𝒓𝒓 − 𝒓𝒓𝑖𝑖 𝑑𝑑𝑑𝑑


𝑉𝑉𝛻𝛻.𝑬𝑬 𝒓𝒓 𝑑𝑑𝑑𝑑


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𝑖𝑖= 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖


=𝑄𝑄𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝜖𝜖0

independent of charge location inside S

independent of shape of surface S


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4 𝜋𝜋 𝜖𝜖0�−∞

+∞

𝜌𝜌 𝒓𝒓𝑞𝒓𝒓 − 𝒓𝒓′

𝒓𝒓 − 𝒓𝒓′ 3𝑑𝑑𝑑𝑑𝑞

Continuous Distribution of Charges 𝜌𝜌 𝒓𝒓𝑞 :

=1

4 𝜋𝜋 𝜖𝜖0�−∞

+∞

𝜌𝜌 𝒓𝒓𝑞 𝛻𝛻−1𝒓𝒓 − 𝒓𝒓𝑞

𝑑𝑑𝑑𝑑𝑞

𝛻𝛻.𝑬𝑬 𝒓𝒓

=1

4 𝜋𝜋 𝜖𝜖0�−∞

+∞

𝜌𝜌 𝒓𝒓𝑞 𝛻𝛻2−1𝒓𝒓 − 𝒓𝒓𝑞


=1

4 𝜋𝜋 𝜖𝜖0�−∞

+∞

𝜌𝜌 𝒓𝒓𝑞 4 𝜋𝜋 𝛿𝛿3 𝒓𝒓 − 𝒓𝒓𝑞 𝑑𝑑𝑑𝑑𝑞 =𝜌𝜌 𝒓𝒓𝜖𝜖0

= 𝛻𝛻.1

4 𝜋𝜋 𝜖𝜖0�−∞

+∞

𝜌𝜌 𝒓𝒓𝑞 𝛻𝛻−1𝒓𝒓 − 𝒓𝒓𝑞



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𝑉𝑉𝛻𝛻.𝑬𝑬 𝒓𝒓 𝑑𝑑𝑑𝑑 = �

𝑉𝑉

𝜌𝜌 𝒓𝒓𝜖𝜖0

𝑑𝑑𝑑𝑑

𝛻𝛻.𝑬𝑬 𝒓𝒓 =𝜌𝜌 𝒓𝒓𝜖𝜖0

Gauss’s Law (differential form)

=𝑄𝑄𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝜖𝜖0

�𝑆𝑆𝑬𝑬 𝒓𝒓 .𝒅𝒅𝒅𝒅 =

𝑄𝑄𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝜖𝜖0

Gauss’s Law (integral form)


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Back to the Electric Field:

= −𝛻𝛻Φ 𝒓𝒓

Φ 𝒓𝒓 ≡1


𝑞𝑞𝑖𝑖𝒓𝒓 − 𝒓𝒓𝑖𝑖

Electric Field 𝑬𝑬 𝒓𝒓is a vector field

Electric Potential Φ 𝒓𝒓is a scalar field

(discrete charges)

=1





21

Back to the Electric field:(continuous charge distribution)

= −𝛻𝛻Φ 𝒓𝒓

Φ 𝒓𝒓 ≡1

4 𝜋𝜋 𝜖𝜖0�−∞

+∞𝜌𝜌 𝒓𝒓𝑞𝒓𝒓 − 𝒓𝒓′


=1

4 𝜋𝜋 𝜖𝜖0�−∞

+∞

𝜌𝜌 𝒓𝒓𝑞 𝛻𝛻−1

𝒓𝒓 − 𝒓𝒓′𝑑𝑑𝑑𝑑𝑞


4 𝜋𝜋 𝜖𝜖0�−∞

+∞

𝜌𝜌 𝒓𝒓𝑞𝒓𝒓 − 𝒓𝒓′

𝒓𝒓 − 𝒓𝒓′ 3𝑑𝑑𝑑𝑑𝑞


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𝑬𝑬 𝒓𝒓 = −𝛻𝛻Φ 𝒓𝒓

Φ𝑞 𝒓𝒓 = Φ 𝒓𝒓 + Λ

𝛻𝛻.𝑬𝑬 𝒓𝒓 =𝜌𝜌 𝒓𝒓𝜖𝜖0

A couple of observations:

(1) Adding a constant Λ to the electric potential Φ 𝒓𝒓 has no impact on the electric field 𝑬𝑬 𝒓𝒓 :

(2) We can solve the electric potential first and then use it to determine the electric field:

𝛻𝛻. −𝛻𝛻Φ 𝒓𝒓 =𝜌𝜌 𝒓𝒓𝜖𝜖0

−𝛻𝛻2Φ 𝒓𝒓 =𝜌𝜌 𝒓𝒓𝜖𝜖0

Poisson’s equation


23

𝛻𝛻 × 𝛻𝛻𝑓𝑓 𝒓𝒓 = 0HW:

𝑬𝑬 𝒓𝒓 = −𝛻𝛻Φ 𝒓𝒓

𝛻𝛻 × 𝑬𝑬 𝒓𝒓

𝛻𝛻 × 𝑬𝑬 𝒓𝒓 = 0


= 𝛻𝛻 × −𝛻𝛻Φ 𝒓𝒓 = −𝛻𝛻 × 𝛻𝛻Φ 𝒓𝒓 = 0

24

Stokes’ Theorem

�𝐶𝐶𝑸𝑸 𝒓𝒓 . 𝒅𝒅𝒓𝒓 = �

𝑆𝑆𝛻𝛻 × 𝑸𝑸 𝒓𝒓 . 𝒅𝒅𝒅𝒅


CS

𝛻𝛻 × 𝑸𝑸 𝒓𝒓

𝑸𝑸 𝒓𝒓

𝐶𝐶 𝐶𝐶

𝑆𝑆𝑆𝑆

Consider a loop C surrounding a surface S

and a vector field 𝑸𝑸 𝒓𝒓 ,

if 𝑸𝑸 𝒓𝒓 is a continuously differentiable vector field:

25

Illustration: Net Loop in the x-y plane


= 𝑸𝑸 𝒓𝒓 . + �𝒆𝒆𝒙𝒙 ∆𝐹𝐹 + 𝑸𝑸 𝒓𝒓 − ∆𝑦𝑦 �𝒆𝒆𝒚𝒚 . − �𝒆𝒆𝒙𝒙 ∆𝐹𝐹

𝐹𝐹

𝑦𝑦

𝑸𝑸 𝒓𝒓 . 𝒅𝒅𝒓𝒓

= 𝑄𝑄𝑥𝑥 𝒓𝒓 − 𝑄𝑄𝑥𝑥 𝒓𝒓 − ∆𝑦𝑦 �𝒆𝒆𝒚𝒚 ∆𝐹𝐹

∆𝐹𝐹

=𝜕𝜕𝑄𝑄𝑥𝑥𝜕𝜕𝑦𝑦 −

𝜕𝜕𝑄𝑄𝑦𝑦𝜕𝜕𝐹𝐹 ∆𝐹𝐹 ∆𝑦𝑦

= 𝑄𝑄𝑥𝑥 𝒓𝒓 − 𝑄𝑄𝑥𝑥 𝒓𝒓 +𝜕𝜕𝑄𝑄𝑥𝑥𝜕𝜕𝑦𝑦 ∆𝑦𝑦 ∆𝐹𝐹

∆𝑦𝑦

+�𝒆𝒆𝒙𝒙

−�𝒆𝒆𝒙𝒙

+ 𝑸𝑸 𝒓𝒓 . + �𝒆𝒆𝒚𝒚 ∆𝑦𝑦 + 𝑸𝑸 𝒓𝒓 + ∆𝐹𝐹 �𝒆𝒆𝒙𝒙 . − �𝒆𝒆𝒚𝒚 ∆𝑦𝑦

+�𝒆𝒆𝒚𝒚 −�𝒆𝒆𝒚𝒚 + 𝑄𝑄𝑦𝑦 𝒓𝒓 − 𝑄𝑄𝑦𝑦 𝒓𝒓 + ∆𝐹𝐹 �𝒆𝒆𝒙𝒙 ∆𝑦𝑦

+ 𝑄𝑄𝑦𝑦 𝒓𝒓 − 𝑄𝑄𝑦𝑦 𝒓𝒓 −𝜕𝜕𝑄𝑄𝑦𝑦𝜕𝜕𝐹𝐹 ∆𝐹𝐹 ∆𝑦𝑦

= 𝛻𝛻 × 𝑸𝑸 𝒛𝒛 ∆𝐹𝐹 ∆𝑦𝑦

26

Net Loop


𝑸𝑸 𝒓𝒓 . 𝒅𝒅𝒓𝒓

= 𝛻𝛻 × 𝑸𝑸 𝑧𝑧 ∆𝐹𝐹 ∆𝑦𝑦 + 𝛻𝛻 × 𝑸𝑸 𝑥𝑥 ∆𝑦𝑦 ∆𝑧𝑧+ 𝛻𝛻 × 𝑸𝑸 𝑦𝑦 ∆𝑧𝑧 ∆𝐹𝐹

= 𝛻𝛻 × 𝑸𝑸 . 𝒅𝒅𝒅𝒅

27

�𝒂𝒂, 𝐶𝐶1

𝒃𝒃𝑬𝑬 𝒓𝒓 . 𝒅𝒅𝒓𝒓 = �

𝐶𝐶𝑬𝑬 𝒓𝒓 . 𝒅𝒅𝒓𝒓+�

𝒃𝒃, 𝐶𝐶2

𝒂𝒂𝑬𝑬 𝒓𝒓 . 𝒅𝒅𝒓𝒓

= �𝑆𝑆𝛻𝛻 × 𝑬𝑬 𝒓𝒓 . 𝒅𝒅𝒅𝒅

= �𝑆𝑆

0 . 𝒅𝒅𝒅𝒅

= 0

�𝒂𝒂, 𝐶𝐶1

𝒃𝒃𝑬𝑬 𝒓𝒓 . 𝒅𝒅𝒓𝒓 = −�

𝒃𝒃, 𝐶𝐶2

𝒂𝒂𝑬𝑬 𝒓𝒓 . 𝒅𝒅𝒓𝒓

= �𝒂𝒂, 𝐶𝐶2

𝒃𝒃𝑬𝑬 𝒓𝒓 . 𝒅𝒅𝒓𝒓 = constant, regardless of the path

𝒂𝒂

𝒃𝒃

𝐶𝐶1 𝐶𝐶2


𝑬𝑬 𝒓𝒓

𝒅𝒅𝒓𝒓

28

= Φ 𝒓𝒓 −Φ −∞

𝑬𝑬 𝒓𝒓 ≡ −𝛻𝛻Φ 𝒓𝒓


= �−∞

𝒓𝒓𝛻𝛻Φ 𝒓𝒓 . 𝒅𝒅𝒓𝒓

= �−∞

𝒓𝒓 𝜕𝜕Φ𝜕𝜕𝐹𝐹

�𝒆𝒆𝑥𝑥 +𝜕𝜕Φ𝜕𝜕𝑦𝑦

�𝒆𝒆𝑦𝑦𝜕𝜕Φ𝜕𝜕𝑧𝑧

�𝒆𝒆𝑧𝑧 . 𝑑𝑑𝐹𝐹 �𝒆𝒆𝑥𝑥 + 𝑑𝑑𝑦𝑦 �𝒆𝒆𝑦𝑦 + 𝑑𝑑𝑧𝑧 �𝒆𝒆𝑧𝑧

= �−∞

𝒓𝒓 𝜕𝜕Φ𝜕𝜕𝐹𝐹

𝑑𝑑𝐹𝐹 +𝜕𝜕Φ𝜕𝜕𝑦𝑦

𝑑𝑑𝑦𝑦 +𝜕𝜕Φ𝜕𝜕𝑧𝑧

𝑑𝑑𝑧𝑧 = �−∞

𝒓𝒓𝑑𝑑Φ

−�−∞

𝒓𝒓𝑬𝑬 𝒓𝒓 . 𝒅𝒅𝒓𝒓

𝒂𝒂 → −∞𝒃𝒃 → 𝒓𝒓

29

Φ 𝒓𝒓 ≡ −�−∞

𝒓𝒓𝑬𝑬 𝒓𝒓𝑞 . 𝒅𝒅𝒓𝒓𝑞

�𝒂𝒂

𝒃𝒃𝑬𝑬 𝒓𝒓𝑞 . 𝒅𝒅𝒓𝒓𝑞 = �

𝒂𝒂

−∞𝑬𝑬 𝒓𝒓𝑞 . 𝒅𝒅𝒓𝒓𝑞 + �

−∞

𝒃𝒃𝑬𝑬 𝒓𝒓𝑞 . 𝒅𝒅𝒓𝒓𝑞

= Φ 𝒂𝒂 −Φ 𝒃𝒃

regardless of the path 𝐶𝐶𝒂𝒂

𝒃𝒃

𝐶𝐶


= −�−∞

𝒂𝒂𝑬𝑬 𝒓𝒓𝑞 . 𝒅𝒅𝒓𝒓𝑞 + �

−∞

𝒃𝒃𝑬𝑬 𝒓𝒓𝑞 . 𝒅𝒅𝒓𝒓𝑞

30

Workdone by the electrostatic force 𝑭𝑭𝑖𝑖𝑎𝑎𝑒𝑒𝑒𝑒𝑖𝑖𝑒𝑒 𝒓𝒓 when a charge q moves from a to b

𝑊𝑊 = �𝒂𝒂

𝒃𝒃𝑭𝑭𝑖𝑖𝑎𝑎𝑒𝑒𝑒𝑒𝑖𝑖𝑒𝑒 𝒓𝒓 . 𝒅𝒅𝒓𝒓

𝑭𝑭𝑖𝑖𝑎𝑎𝑒𝑒𝑒𝑒𝑖𝑖𝑒𝑒 𝒓𝒓 = 𝑞𝑞 𝑬𝑬 𝒓𝒓

𝑊𝑊 = �𝒂𝒂

𝒃𝒃𝑞𝑞 𝑬𝑬 𝒓𝒓 . 𝒅𝒅𝒓𝒓 = 𝑞𝑞 Φ 𝒂𝒂 − Φ 𝒃𝒃

𝒂𝒂

𝒃𝒃

𝐶𝐶

regardless of the path


31

External Workdone by an external force to perfectly balance the electrostatic force

𝑊𝑊𝑖𝑖𝑥𝑥𝑒𝑒 = 𝑞𝑞 Φ 𝒃𝒃 − Φ 𝒂𝒂𝒂𝒂

𝒃𝒃

𝑊𝑊𝑖𝑖𝑥𝑥𝑒𝑒 = 𝑞𝑞 Φ 𝒓𝒓 − Φ −∞

= 𝑞𝑞 Φ 𝒓𝒓

𝑭𝑭𝑖𝑖𝑥𝑥𝑒𝑒 𝒓𝒓 = − 𝑭𝑭𝑖𝑖𝑎𝑎𝑒𝑒𝑒𝑒𝑖𝑖𝑒𝑒 𝒓𝒓

stored energy !!Prof. Sergio B. MendesSpring 2018

𝐶𝐶

−∞

𝒓𝒓

𝐶𝐶

𝒂𝒂 → −∞0


Recapping:




= −𝛻𝛻Φ 𝒓𝒓

Φ 𝒓𝒓 ≡1


𝑞𝑞𝑖𝑖𝒓𝒓 − 𝒓𝒓𝑖𝑖

=1





Recapping, continued:𝑬𝑬 𝒓𝒓 = −𝛻𝛻Φ 𝒓𝒓

𝛻𝛻 × 𝑬𝑬 𝒓𝒓 = −𝛻𝛻 × 𝛻𝛻Φ 𝒓𝒓 = 0

�𝐶𝐶𝑬𝑬 𝒓𝒓 . 𝒅𝒅𝒓𝒓 = �

𝑆𝑆𝛻𝛻 × 𝑬𝑬 𝒓𝒓 . 𝒅𝒅𝒅𝒅 = 0

−�−∞

𝒓𝒓𝑬𝑬 𝒓𝒓 . 𝒅𝒅𝒓𝒓 = Φ 𝒓𝒓 −Φ −∞


Recapping, continued:

−�−∞

𝒓𝒓𝑬𝑬 𝒓𝒓 . 𝒅𝒅𝒓𝒓 = Φ 𝒓𝒓

𝑭𝑭𝑖𝑖𝑥𝑥𝑒𝑒 𝒓𝒓 = − 𝑭𝑭 𝒓𝒓

𝑊𝑊𝑖𝑖𝑥𝑥𝑒𝑒 = �−∞

𝒓𝒓𝑭𝑭𝑖𝑖𝑥𝑥𝑒𝑒 𝒓𝒓 . 𝒅𝒅𝒓𝒓 = −�

−∞

𝒓𝒓𝑞𝑞 𝑬𝑬 𝒓𝒓 . 𝒅𝒅𝒓𝒓

= − 𝑞𝑞 𝑬𝑬 𝒓𝒓

= 𝑞𝑞 Φ 𝒓𝒓

35

𝑊𝑊1 & 2 = 𝑞𝑞1 Φ2 𝒓𝒓1

= 𝑞𝑞2 Φ1 𝒓𝒓2

=12𝑞𝑞1 Φ2 𝒓𝒓1 +

12𝑞𝑞2 Φ1 𝒓𝒓2

𝑞𝑞1 𝑞𝑞2

𝒪𝒪

𝒓𝒓1 𝒓𝒓2

External work done to assemble charges = Stored Energy

two point charges


𝑞𝑞2

36

𝑞𝑞1 𝑞𝑞2

𝒪𝒪

𝒓𝒓1 𝒓𝒓2

Additional work for bringing a third charge

𝑞𝑞3∆𝑊𝑊 = 𝑞𝑞3 Φ1 𝒓𝒓3 + 𝑞𝑞3 Φ2 𝒓𝒓3

= 𝑞𝑞1 Φ3 𝒓𝒓1 + 𝑞𝑞2 Φ3 𝒓𝒓2

=12𝑞𝑞3 Φ1 𝒓𝒓3 +

12𝑞𝑞1 Φ3 𝒓𝒓1 +

12𝑞𝑞3 Φ2 𝒓𝒓3 +



𝒓𝒓3

𝑞𝑞1 𝑞𝑞2

37

𝑊𝑊1 & 2 & 3 =12𝑞𝑞1 Φ2 𝒓𝒓1 +


+12𝑞𝑞3 Φ1 𝒓𝒓3 +


+12𝑞𝑞3 Φ2 𝒓𝒓3 +


=12𝑞𝑞1 Φ2 𝒓𝒓1 + Φ3 𝒓𝒓1

+12𝑞𝑞2 Φ3 𝒓𝒓2 + Φ1 𝒓𝒓2

+12𝑞𝑞3 Φ2 𝒓𝒓3 + Φ3 𝒓𝒓3

Work needed to assemble three charges(stored energy)


38

𝑊𝑊 = �𝑖𝑖, 𝑗𝑗≠𝑖𝑖

𝑁𝑁12𝑞𝑞𝑖𝑖 Φ𝑗𝑗 𝒓𝒓𝑖𝑖

Work needed to assemble “N” charges(stored energy)


39

𝑑𝑑𝑊𝑊 =12𝑑𝑑𝑞𝑞 𝒓𝒓 Φ 𝒓𝒓

𝑑𝑑𝑞𝑞 𝒓𝒓 = 𝜌𝜌 𝒓𝒓 𝑑𝑑𝑑𝑑

𝜌𝜌 𝒓𝒓 = − 𝜖𝜖0 𝛻𝛻2Φ 𝒓𝒓

= −12𝜖𝜖0 Φ 𝒓𝒓 𝛻𝛻2Φ 𝒓𝒓 𝑑𝑑𝑑𝑑

Energy Stored in a Continuous Distribution of Charges

−Φ 𝒓𝒓 𝛻𝛻2Φ 𝒓𝒓 = − 𝛻𝛻. Φ 𝒓𝒓 𝛻𝛻Φ 𝒓𝒓 + 𝛻𝛻Φ 𝒓𝒓 . 𝛻𝛻Φ 𝒓𝒓HW:


= − 𝜖𝜖0 𝛻𝛻2Φ 𝒓𝒓 𝑑𝑑𝑑𝑑

40

=12𝜖𝜖0 − 𝛻𝛻. Φ 𝒓𝒓 𝛻𝛻Φ 𝒓𝒓 + 𝛻𝛻Φ 𝒓𝒓 . 𝛻𝛻Φ 𝒓𝒓 𝑑𝑑𝑑𝑑

𝑊𝑊 =12𝜖𝜖0�

−∞

+∞

− 𝛻𝛻. Φ 𝒓𝒓 𝛻𝛻Φ 𝒓𝒓 + 𝛻𝛻Φ 𝒓𝒓 . 𝛻𝛻Φ 𝒓𝒓 𝑑𝑑𝑑𝑑

=12𝜖𝜖0 − �

−∞

+∞

Φ 𝒓𝒓 𝛻𝛻Φ 𝒓𝒓 . 𝒅𝒅𝒅𝒅 + �−∞

+∞

𝛻𝛻Φ 𝒓𝒓 . 𝛻𝛻Φ 𝒓𝒓 𝑑𝑑𝑑𝑑

=12𝜖𝜖0�

−∞

+∞

𝛻𝛻Φ 𝒓𝒓 2 𝑑𝑑𝑑𝑑


𝑑𝑑𝑊𝑊 = −12𝜖𝜖0 Φ 𝒓𝒓 𝛻𝛻2Φ 𝒓𝒓 𝑑𝑑𝑑𝑑

0

=12𝜖𝜖0�

−∞

+∞

𝑬𝑬 𝒓𝒓 2 𝑑𝑑𝑑𝑑

41

𝐹𝐹𝐸𝐸 𝒓𝒓 ≡𝑑𝑑𝑊𝑊𝑑𝑑𝑑𝑑

=12𝜖𝜖0 𝑬𝑬 𝒓𝒓 2

(Electric) Energy and Energy Density


𝑊𝑊 =12𝜖𝜖0�

−∞

+∞

𝑬𝑬 𝒓𝒓 2 𝑑𝑑𝑑𝑑

Magnetostatic Theory

42

steady (constant) current

current density: 𝑱𝑱 𝒓𝒓, 𝑡𝑡 = 𝑱𝑱 𝒓𝒓

𝑱𝑱 ≡𝑑𝑑𝑑𝑑𝑑𝑑𝑆𝑆

�𝒏𝒏


43

Conservation of Electric Charge

�s𝑱𝑱 𝒓𝒓, 𝑡𝑡 . 𝒅𝒅𝒅𝒅 = −

𝑑𝑑𝑄𝑄𝑉𝑉𝑑𝑑𝑡𝑡

�𝑉𝑉𝛁𝛁. 𝑱𝑱 𝒓𝒓, 𝑡𝑡 𝑑𝑑𝑑𝑑

𝛁𝛁. 𝑱𝑱 𝒓𝒓, 𝑡𝑡 = −𝜕𝜕𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

𝜌𝜌 𝒓𝒓, 𝑡𝑡 = 𝜌𝜌 𝒓𝒓 𝜕𝜕𝜌𝜌 𝒓𝒓𝜕𝜕𝑡𝑡

= 0 𝛁𝛁. 𝑱𝑱 𝒓𝒓, 𝑡𝑡 = 𝛁𝛁. 𝑱𝑱 𝒓𝒓 = 0if then and

always valid due to conservation of charge


= −𝑑𝑑𝑑𝑑𝑡𝑡�

𝑉𝑉𝜌𝜌 𝒓𝒓, 𝑡𝑡 𝑑𝑑𝑑𝑑

−�𝑉𝑉

𝜕𝜕𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

𝑑𝑑𝑑𝑑=

44

𝛁𝛁. 𝑱𝑱 𝒓𝒓 = 0

Satisfies × ViolatesProf. Sergio B. MendesSpring 2018

�s𝑱𝑱 𝒓𝒓 . 𝒅𝒅𝒅𝒅 = 0

�𝑉𝑉𝛁𝛁. 𝑱𝑱 𝒓𝒓, 𝑡𝑡 𝑑𝑑𝑑𝑑 = 0

Continuous Flow of Electric Current

45

Milestones in Magnetostatics• 2637 B.C.: reports of magnets by Chinese civilization

• 3rd century A.D.: Chinese ships used compass

• 1600: Gilbert, (De Magnete) described Earth as a magnet

• 800 A.D.: Greek reports on lodestones (magnetite Fe2O3)

• 1820: Oersted, an electric current deflects a compass

• 1830’s: Ampere, nearby currents create a force on each other


46

Magnetic Force• On a moving 𝒗𝒗 electric charge (𝑞𝑞) due to a magnetic field (𝑩𝑩)

𝑑𝑑𝑭𝑭 = 𝑑𝑑𝑞𝑞 𝒗𝒗 × 𝑩𝑩

𝑭𝑭 = 𝑞𝑞 𝒗𝒗 × 𝑩𝑩 Lorentz force


• On an electric current due to a magnetic field 𝑩𝑩

= I dr × 𝑩𝑩 = dV 𝑱𝑱 × 𝑩𝑩

• For a non-zero magnetic force, charges must be moving and in a path not collinear with the magnetic field

• [B] = Tesla in S.I. units

• Static magnetic force is always orthogonal to the instantaneous direction of charge motion

• Work: 𝑑𝑑𝑊𝑊 = 𝑭𝑭 . 𝒅𝒅𝒓𝒓 = 𝑞𝑞 𝒗𝒗 × 𝑩𝑩 . 𝒗𝒗 𝑑𝑑𝑡𝑡 = 0

47

Magnetic Fieldis created by electric charges in motion (electric current)

𝑩𝑩 𝒓𝒓 =𝜇𝜇𝑜𝑜4 𝜋𝜋

�𝐶𝐶

I 𝒓𝒓′ d 𝒓𝒓′ ×𝒓𝒓 − 𝒓𝒓′


=𝜇𝜇𝑜𝑜4 𝜋𝜋

�−∞

+∞

𝑱𝑱 𝒓𝒓′ dV𝑞 ×𝒓𝒓 − 𝒓𝒓′


Biot-Savart Law

𝜇𝜇𝑜𝑜 = 4 π × 10−7𝑁𝑁𝐴𝐴2

(permeability of free space)Prof. Sergio B. MendesSpring 2018

48

𝑩𝑩 𝒓𝒓 =𝜇𝜇𝑜𝑜4 𝜋𝜋

�−∞

+∞

dV𝑞 𝑱𝑱 𝒓𝒓′ ×𝒓𝒓 − 𝒓𝒓′


𝛻𝛻. 𝒂𝒂 × 𝒃𝒃 = 𝒃𝒃 .𝛻𝛻 × 𝒂𝒂 − 𝒂𝒂 . 𝛻𝛻 × 𝒃𝒃HW:

=𝜇𝜇𝑜𝑜4 𝜋𝜋

�−∞

+∞

dV𝑞𝒓𝒓 − 𝒓𝒓′

𝒓𝒓 − 𝒓𝒓𝑞 3.𝛻𝛻 × 𝑱𝑱 𝒓𝒓′ − 𝑱𝑱 𝒓𝒓′ .𝛻𝛻 ×

𝒓𝒓 − 𝒓𝒓′


= −𝜇𝜇𝑜𝑜4 𝜋𝜋

�−∞

+∞

dV𝑞 𝑱𝑱 𝒓𝒓′ .𝛻𝛻 × 𝛻𝛻−1𝒓𝒓 − 𝒓𝒓𝑞

𝛻𝛻 × 𝛻𝛻𝑓𝑓 𝒓𝒓 = 𝟎𝟎 𝛻𝛻.𝑩𝑩 𝒓𝒓 = 𝟎𝟎because: = 0,


0

𝛻𝛻. 𝛻𝛻.

49

𝛻𝛻 × 𝑩𝑩 𝒓𝒓 = 𝛻𝛻 ×𝜇𝜇𝑜𝑜4 𝜋𝜋

�−∞

+∞

dV𝑞 J 𝒓𝒓′ ×𝒓𝒓 − 𝒓𝒓′

𝒓𝒓 − 𝒓𝒓𝑞 3= 𝜇𝜇𝑜𝑜 J 𝒓𝒓HW:

𝛻𝛻 × 𝑩𝑩 𝒓𝒓 = 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓Ampère’s Law(differential form)

�𝑆𝑆𝛻𝛻 × 𝑩𝑩 𝒓𝒓 .𝒅𝒅𝒅𝒅 = 𝜇𝜇𝑜𝑜�

𝑆𝑆𝑱𝑱 𝒓𝒓 .𝒅𝒅𝒅𝒅

�𝐶𝐶𝑩𝑩 𝒓𝒓 .𝒅𝒅𝒓𝒓 = 𝜇𝜇𝑜𝑜 𝑑𝑑

Ampère’s Law(integral form)


50

Vector Potential: 𝑨𝑨 𝒓𝒓

𝛻𝛻.𝑩𝑩 𝒓𝒓 = 0

𝑩𝑩 𝒓𝒓 ≡ 𝛻𝛻 × 𝑨𝑨 𝒓𝒓

HW: 𝛻𝛻. 𝛻𝛻 × 𝑸𝑸 = 0

𝛻𝛻.𝑩𝑩 𝒓𝒓 = 𝛻𝛻. 𝛻𝛻 × 𝑨𝑨 𝒓𝒓


= 0

51

𝛻𝛻 × 𝛻𝛻 × 𝑨𝑨 𝒓𝒓 = 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓

𝛻𝛻 × 𝛻𝛻 × 𝒂𝒂 = 𝛻𝛻 𝛻𝛻.𝒂𝒂 − 𝛻𝛻𝟐𝟐𝒂𝒂HW:


𝛻𝛻 × 𝑩𝑩 𝒓𝒓 = 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓

𝜵𝜵 × 𝜵𝜵 × 𝑨𝑨 𝒓𝒓 = 𝜵𝜵 𝜵𝜵.𝑨𝑨 𝒓𝒓 − 𝛻𝛻𝟐𝟐𝑨𝑨 𝒓𝒓 = 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓

𝑩𝑩 𝒓𝒓 ≡ 𝜵𝜵 × 𝑨𝑨 𝒓𝒓


𝑨𝑨𝑞 𝒓𝒓 = 𝑨𝑨 𝒓𝒓 + 𝜵𝜵𝛬𝛬 𝒓𝒓

𝜵𝜵 × 𝑨𝑨′ 𝒓𝒓 = 𝜵𝜵 × 𝑨𝑨 𝒓𝒓

𝛁𝛁 × 𝛁𝛁𝛬𝛬 𝒓𝒓 = 0

because

then

We have certain freedom to choose the Vector Potential:

= 𝑩𝑩 𝒓𝒓


𝑨𝑨𝑞 𝒓𝒓 = 𝑨𝑨 𝒓𝒓 + 𝜵𝜵𝛬𝛬 𝒓𝒓

𝜵𝜵.𝑨𝑨𝑞 𝒓𝒓 = 𝜵𝜵.𝑨𝑨 𝒓𝒓 + 𝜵𝜵.𝛻𝛻𝛬𝛬 𝒓𝒓

= 𝜵𝜵.𝑨𝑨 𝒓𝒓 + 𝛻𝛻2𝛬𝛬 𝒓𝒓

Now, we will find 𝛬𝛬 𝒓𝒓 such that 𝛻𝛻2𝛬𝛬 𝒓𝒓 = −𝛁𝛁.𝑨𝑨 𝒓𝒓 .

𝛁𝛁.𝑨𝑨𝑞 𝒓𝒓 = 0

𝜵𝜵 × 𝜵𝜵 × 𝑨𝑨𝑞 𝒓𝒓 = 𝜵𝜵 𝜵𝜵.𝑨𝑨𝑞 𝒓𝒓 − 𝛻𝛻𝟐𝟐𝑨𝑨𝑞 𝒓𝒓 = 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓

−𝛻𝛻𝟐𝟐𝑨𝑨𝑞 𝒓𝒓 = 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓

Coulomb gauge

How can we benefit from this freedom?

0

Then we will get:


Big picture in

Electrostatics and Magnetostatics:


𝑩𝑩 𝒓𝒓 = 𝜵𝜵 × 𝑨𝑨 𝒓𝒓

−𝛻𝛻𝟐𝟐𝑨𝑨 𝒓𝒓 = 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓−𝛻𝛻2Φ 𝒓𝒓 =𝜌𝜌 𝒓𝒓𝜖𝜖0

𝑬𝑬 𝒓𝒓 = −𝜵𝜵Φ 𝒓𝒓

Φ 𝒓𝒓 ≡1

4 𝜋𝜋 𝜖𝜖0�−∞

+∞𝜌𝜌 𝒓𝒓𝑞𝒓𝒓 − 𝒓𝒓′

𝑑𝑑𝑑𝑑𝑞 𝑨𝑨 𝒓𝒓 ≡𝜇𝜇𝑜𝑜

4 𝜋𝜋�−∞

+∞𝑱𝑱 𝒓𝒓𝑞𝒓𝒓 − 𝒓𝒓′


Electrostatics Magnetostatics

𝜌𝜌 𝒓𝒓𝑱𝑱 𝒓𝒓

𝜵𝜵. 𝑱𝑱 𝒓𝒓, 𝑡𝑡 = 0−𝜕𝜕𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

= 𝜵𝜵. 𝑱𝑱 𝒓𝒓, 𝑡𝑡𝜕𝜕𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡 = 0

charge conservation

𝑰𝑰

Electrodynamic Theory

56

𝜌𝜌 𝒓𝒓, 𝑡𝑡


𝑱𝑱 𝒓𝒓, 𝑡𝑡−𝜕𝜕𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

= 𝜵𝜵. 𝑱𝑱 𝒓𝒓, 𝑡𝑡

time-dependent theory for electric and magnetic fields

}


What remains valid ?


𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝑬𝑬, 𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒆𝒆𝒅𝒅 𝑡𝑡 ≡ �𝑆𝑆𝑬𝑬 𝒓𝒓, 𝑡𝑡 .𝒅𝒅𝒅𝒅 = �

𝑉𝑉𝜵𝜵.𝑬𝑬 𝒓𝒓, 𝑡𝑡 𝑑𝑑𝑑𝑑 = �

𝑉𝑉

𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜖𝜖0

𝑑𝑑𝑑𝑑

𝜵𝜵.𝑬𝑬 𝒓𝒓, 𝑡𝑡 =𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜖𝜖0

= �𝑉𝑉𝜵𝜵.𝑬𝑬 𝒓𝒓 𝑑𝑑𝑑𝑑𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝑬𝑬, 𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒆𝒆𝒅𝒅 ≡ �

𝑆𝑆𝑬𝑬 𝒓𝒓 .𝒅𝒅𝒅𝒅 = �

𝑉𝑉

𝜌𝜌 𝒓𝒓𝜖𝜖0

𝑑𝑑𝑑𝑑


𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝑩𝑩, 𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒆𝒆𝒅𝒅 𝑡𝑡 ≡ �𝑆𝑆𝑩𝑩 𝒓𝒓, 𝑡𝑡 .𝒅𝒅𝒅𝒅 = �

𝑉𝑉𝜵𝜵.𝑩𝑩 𝒓𝒓, 𝑡𝑡 𝑑𝑑𝑑𝑑 = 0

𝜵𝜵.𝑩𝑩 𝒓𝒓, 𝑡𝑡 = 0

𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝑩𝑩, 𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒆𝒆𝒅𝒅 ≡ �𝑆𝑆𝑩𝑩 𝒓𝒓 .𝒅𝒅𝒅𝒅 = �

𝑉𝑉𝜵𝜵.𝑩𝑩 𝒓𝒓 𝑑𝑑𝑑𝑑 = 0


What is new ?


𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝑩𝑩 𝑡𝑡 ≡ �𝑆𝑆𝑩𝑩 𝒓𝒓, 𝑡𝑡 . 𝒅𝒅𝒅𝒅

−𝑑𝑑𝑑𝑑𝑡𝑡𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝑩𝑩 𝑡𝑡 = �

𝐶𝐶𝑬𝑬 𝒓𝒓, 𝑡𝑡 . 𝒅𝒅𝒓𝒓

Faraday’s Law

−𝑑𝑑𝑑𝑑𝑡𝑡�𝑆𝑆𝑩𝑩 𝒓𝒓, 𝑡𝑡 . 𝒅𝒅𝒅𝒅

−𝜕𝜕𝑩𝑩 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

= 𝜵𝜵 × 𝑬𝑬 𝒓𝒓, 𝑡𝑡

= �𝑆𝑆𝜵𝜵 × 𝑬𝑬 𝒓𝒓, 𝑡𝑡 . 𝒅𝒅𝒅𝒅


What needs to be modified ?


𝜵𝜵. 𝑱𝑱 𝒓𝒓, 𝑡𝑡 +𝜕𝜕𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

= 0


𝜕𝜕𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

= 𝜖𝜖0𝜕𝜕 𝜵𝜵.𝑬𝑬 𝒓𝒓, 𝑡𝑡

𝜕𝜕𝑡𝑡

𝜵𝜵. 𝑱𝑱 𝒓𝒓, 𝑡𝑡 + 𝜖𝜖0𝜕𝜕 𝜵𝜵.𝑬𝑬 𝒓𝒓, 𝑡𝑡

𝜕𝜕𝑡𝑡= 0

𝜵𝜵. 𝑱𝑱 𝒓𝒓, 𝑡𝑡 + 𝜖𝜖0𝜕𝜕𝑬𝑬 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

= 0

= 𝜵𝜵. 𝜵𝜵 ×𝑩𝑩 𝒓𝒓, 𝑡𝑡𝜇𝜇𝑜𝑜

𝜵𝜵 × 𝑩𝑩 𝒓𝒓, 𝑡𝑡 = 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕𝑬𝑬 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

= 𝜵𝜵. 𝜵𝜵 × 𝑸𝑸 𝒓𝒓, 𝑡𝑡


In summary, Maxwell’s equations:


𝜵𝜵.𝑩𝑩 𝒓𝒓, 𝑡𝑡 = 0

𝜵𝜵 × 𝑬𝑬 𝒓𝒓, 𝑡𝑡 = −𝜕𝜕𝑩𝑩 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡


1.

2.

3.

4.

Gauss’s Law

Faraday’s Law

GeneralizedAmpère’s law

Gauss’s Law of Magnetism

65

Vector Potential: 𝑨𝑨 𝒓𝒓, 𝑡𝑡

𝜵𝜵.𝑩𝑩 𝒓𝒓, 𝒕𝒕 = 0

𝑩𝑩 𝒓𝒓, 𝑡𝑡 ≡ 𝜵𝜵 × 𝑨𝑨 𝒓𝒓, 𝑡𝑡 𝜵𝜵.𝑩𝑩 𝒓𝒓, 𝑡𝑡 = 𝜵𝜵. 𝜵𝜵 × 𝑨𝑨 𝒓𝒓, 𝑡𝑡 = 0



Scalar Potential: Φ 𝒓𝒓, 𝑡𝑡


𝑩𝑩 𝒓𝒓, 𝑡𝑡 = 𝜵𝜵 × 𝑨𝑨 𝒓𝒓, 𝑡𝑡

𝜵𝜵 × 𝑬𝑬 𝒓𝒓, 𝑡𝑡 = −𝜕𝜕 𝜵𝜵 × 𝑨𝑨 𝒓𝒓, 𝑡𝑡

𝜕𝜕𝑡𝑡

𝜵𝜵 × 𝑬𝑬 𝒓𝒓, 𝑡𝑡 +𝜕𝜕𝑨𝑨 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

= 𝟎𝟎

𝑬𝑬 𝒓𝒓, 𝑡𝑡 +𝜕𝜕𝑨𝑨 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

≡ −𝜵𝜵Φ 𝒓𝒓, 𝑡𝑡

𝑬𝑬 𝒓𝒓, 𝑡𝑡 = −𝜵𝜵Φ 𝒓𝒓, 𝑡𝑡 −𝜕𝜕𝑨𝑨 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

&

≡ 𝜵𝜵 × −𝜵𝜵Φ 𝒓𝒓, 𝑡𝑡


𝑩𝑩 𝒓𝒓, 𝒕𝒕 = 𝜵𝜵 × 𝑨𝑨 𝒓𝒓, 𝑡𝑡

Fields: 𝑬𝑬 𝒓𝒓, 𝑡𝑡 & 𝑩𝑩 𝒓𝒓, 𝑡𝑡

&

Potentials: Φ 𝒓𝒓, 𝑡𝑡 & 𝑨𝑨 𝒓𝒓, 𝑡𝑡



How to determine

the scalar Φ 𝒓𝒓, 𝑡𝑡 and vector 𝑨𝑨 𝒓𝒓, 𝑡𝑡 potentials

directly from

the charge 𝜌𝜌 𝒓𝒓, 𝑡𝑡 and current 𝑱𝑱 𝒓𝒓, 𝑡𝑡 densities ?




𝜵𝜵. −𝜵𝜵Φ 𝒓𝒓, 𝑡𝑡 −𝜕𝜕𝑨𝑨 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

=

−𝛻𝛻2Φ 𝒓𝒓, 𝑡𝑡 −𝜕𝜕𝜕𝜕𝑡𝑡

𝜵𝜵.𝑨𝑨 𝒓𝒓, 𝑡𝑡 =𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜖𝜖0

Gauss’s Law



𝜵𝜵 × 𝜵𝜵 × 𝑨𝑨 𝒓𝒓, 𝑡𝑡 − 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕𝜕𝜕𝑡𝑡

−𝜵𝜵Φ 𝒓𝒓, 𝑡𝑡 −𝜕𝜕𝑨𝑨 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

= 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓, 𝑡𝑡

𝑩𝑩 𝒓𝒓, 𝑡𝑡 = 𝜵𝜵 × 𝑨𝑨 𝒓𝒓, 𝑡𝑡

𝜵𝜵 × 𝑩𝑩 𝒓𝒓, 𝑡𝑡 − 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕𝑬𝑬 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡


𝜵𝜵 × 𝜵𝜵 × 𝒂𝒂 = 𝜵𝜵 𝜵𝜵.𝒂𝒂 − 𝛻𝛻𝟐𝟐𝒂𝒂

𝜵𝜵 𝜵𝜵.𝑨𝑨 𝒓𝒓, 𝑡𝑡 − 𝛻𝛻𝟐𝟐𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0 𝜵𝜵𝜕𝜕Φ 𝒓𝒓, 𝑡𝑡

𝜕𝜕𝑡𝑡 +𝜕𝜕2𝑨𝑨 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡2 = 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓, 𝑡𝑡

Remember HW:

−𝛻𝛻𝟐𝟐𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕2𝑨𝑨 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡2 + 𝜵𝜵 𝜵𝜵.𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0

𝜕𝜕Φ 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡 = 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓, 𝑡𝑡

Ampère’s Law


−𝛻𝛻2Φ 𝒓𝒓, 𝑡𝑡 −𝜕𝜕𝜕𝜕𝑡𝑡

𝛻𝛻.𝑨𝑨 𝒓𝒓, 𝑡𝑡 =𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜖𝜖0

−𝛻𝛻𝟐𝟐𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕2𝑨𝑨 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡2 + 𝛻𝛻 𝛻𝛻.𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0

𝜕𝜕𝜕𝜕𝑡𝑡 Φ 𝒓𝒓, 𝑡𝑡 = 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓, 𝑡𝑡

The scalar Φ 𝒓𝒓, 𝑡𝑡 and vector 𝑨𝑨 𝒓𝒓, 𝑡𝑡 potentials can then be determined from

the charge 𝜌𝜌 𝒓𝒓, 𝑡𝑡 and current 𝑱𝑱 𝒓𝒓, 𝑡𝑡 densities:


How can we simplify those equations ?


We have certain freedom to choose the vector and scalar potentials:

𝑨𝑨𝑞 𝒓𝒓, 𝑡𝑡 ≡ 𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜵𝜵𝛬𝛬 𝒓𝒓, 𝑡𝑡

Φ′ 𝒓𝒓, 𝑡𝑡 ≡ Φ 𝒓𝒓, 𝑡𝑡 −𝜕𝜕𝛬𝛬 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

𝜵𝜵 × 𝑨𝑨𝑞 𝒓𝒓, 𝑡𝑡

−𝜵𝜵Φ𝑞 𝒓𝒓, 𝑡𝑡 −𝜕𝜕𝑨𝑨𝑞 𝒓𝒓, 𝑡𝑡

𝜕𝜕𝑡𝑡

= −𝜵𝜵Φ 𝒓𝒓, 𝑡𝑡 −𝜕𝜕𝜕𝜕𝑡𝑡𝑨𝑨 𝒓𝒓, 𝑡𝑡

= −𝜵𝜵 Φ 𝒓𝒓, 𝑡𝑡 −𝜕𝜕𝛬𝛬 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

= 𝑬𝑬 𝒓𝒓, 𝑡𝑡

= 𝑩𝑩 𝒓𝒓, 𝑡𝑡

Conclusion: if 𝑨𝑨 𝒓𝒓, 𝑡𝑡 & Φ 𝒓𝒓, 𝑡𝑡 is a solution then 𝑨𝑨𝑞 𝒓𝒓, 𝑡𝑡 & Φ′ 𝒓𝒓, 𝑡𝑡 (as defined above) is also a

solution, and vice-versa.

= 𝜵𝜵 × 𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜵𝜵𝛬𝛬 𝒓𝒓, 𝑡𝑡 = 𝜵𝜵 × 𝑨𝑨 𝒓𝒓, 𝑡𝑡

−𝜕𝜕𝜕𝜕𝑡𝑡 𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜵𝜵𝛬𝛬 𝒓𝒓, 𝑡𝑡

??


𝑨𝑨𝑞 𝒓𝒓, 𝑡𝑡 = 𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜵𝜵𝛬𝛬 𝒓𝒓, 𝑡𝑡

Φ′ 𝒓𝒓, 𝑡𝑡 = Φ 𝒓𝒓, 𝑡𝑡 −𝜕𝜕𝛬𝛬 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

How can we use the freedom in the choice of the potentials

−𝛻𝛻2Φ𝑞 𝒓𝒓, 𝑡𝑡 −𝜕𝜕𝜕𝜕𝑡𝑡

𝜵𝜵.𝑨𝑨𝑞 𝒓𝒓, 𝑡𝑡 =𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜖𝜖0

−𝛻𝛻𝟐𝟐𝑨𝑨𝑞 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕2𝑨𝑨𝑞 𝒓𝒓, 𝑡𝑡

𝜕𝜕𝑡𝑡2+ 𝜵𝜵 𝜵𝜵.𝑨𝑨𝑞 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0

𝜕𝜕𝜕𝜕𝑡𝑡Φ𝑞 𝒓𝒓, 𝑡𝑡 = 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓, 𝑡𝑡

to simplify the following equations:


𝑨𝑨𝑞 𝒓𝒓, 𝑡𝑡 = 𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜵𝜵𝛬𝛬 𝒓𝒓, 𝑡𝑡

Φ′ 𝒓𝒓, 𝑡𝑡 = Φ 𝒓𝒓, 𝑡𝑡 −𝜕𝜕𝛬𝛬 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

𝜵𝜵.𝑨𝑨𝑞 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕𝜕𝜕𝑡𝑡Φ𝑞 𝒓𝒓, 𝑡𝑡 = 𝜵𝜵.𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜵𝜵.𝜵𝜵𝛬𝛬 𝒓𝒓, 𝑡𝑡 +

+ 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕𝜕𝜕𝑡𝑡Φ 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0

𝜕𝜕𝜕𝜕𝑡𝑡

−𝜕𝜕𝛬𝛬 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

𝛻𝛻𝟐𝟐𝛬𝛬 𝒓𝒓, 𝑡𝑡 − 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕2𝛬𝛬 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡2

= − 𝜵𝜵.𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕𝜕𝜕𝑡𝑡Φ 𝒓𝒓, 𝑡𝑡

We now solve for 𝛬𝛬 𝒓𝒓, 𝑡𝑡 such that 𝜵𝜵.𝑨𝑨𝑞 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕𝜕𝜕𝑒𝑒Φ𝑞 𝒓𝒓, 𝑡𝑡 = 𝟎𝟎 :

= 𝑓𝑓 𝒓𝒓, 𝑡𝑡





𝜵𝜵.𝑨𝑨′ 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕𝜕𝜕𝑡𝑡Φ′ 𝒓𝒓, 𝑡𝑡 = 0

−𝛻𝛻2Φ𝑞 𝒓𝒓, 𝑡𝑡 −𝜕𝜕𝜕𝜕𝑡𝑡

𝜵𝜵.𝑨𝑨𝑞 𝒓𝒓, 𝑡𝑡 =𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜖𝜖0




−𝛻𝛻𝟐𝟐𝑨𝑨′ 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕2𝑨𝑨′ 𝒓𝒓, 𝑡𝑡

𝜕𝜕𝑡𝑡2= 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓, 𝑡𝑡

−𝛻𝛻2Φ′ 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕2Φ𝑞 𝒓𝒓, 𝑡𝑡

𝜕𝜕𝑡𝑡2=𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜖𝜖0

Lorenz gauge

= 0


𝛻𝛻.𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕𝜕𝜕𝑡𝑡Φ 𝒓𝒓, 𝑡𝑡 = 0

−𝛻𝛻𝟐𝟐𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕2𝑨𝑨 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡2


−𝛻𝛻2Φ 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕2Φ 𝒓𝒓, 𝑡𝑡


In the Lorenz gauge:

𝜌𝜌 𝒓𝒓, 𝑡𝑡 determines Φ 𝒓𝒓, 𝑡𝑡

𝑱𝑱 𝒓𝒓, 𝑡𝑡 determines 𝑨𝑨 𝒓𝒓, 𝑡𝑡


Recapping:

𝜵𝜵. 𝑱𝑱 𝒓𝒓, 𝑡𝑡 +𝜕𝜕𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

= 0


𝜵𝜵.𝑩𝑩 𝒓𝒓, 𝑡𝑡 = 0



𝑭𝑭 = 𝑞𝑞 𝑬𝑬 + 𝒗𝒗 × 𝑩𝑩


Recapping, cont.:



−𝛻𝛻2Φ 𝒓𝒓, 𝑡𝑡 −𝜕𝜕𝜕𝜕𝑡𝑡



+ 𝜵𝜵 𝜵𝜵.𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕Φ 𝒓𝒓, 𝑡𝑡

𝜕𝜕𝑡𝑡= 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓, 𝑡𝑡


Recapping, cont.:



Φ′ 𝒓𝒓, 𝑡𝑡 ≡ Φ 𝒓𝒓, 𝑡𝑡 −𝜕𝜕𝛬𝛬 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

𝑨𝑨𝑞 𝒓𝒓, 𝑡𝑡 ≡ 𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜵𝜵𝛬𝛬 𝒓𝒓, 𝑡𝑡

= 𝜵𝜵 × 𝑨𝑨𝑞 𝒓𝒓, 𝑡𝑡

= −𝜵𝜵Φ𝑞 𝒓𝒓, 𝑡𝑡 −𝜕𝜕𝑨𝑨𝑞 𝒓𝒓, 𝑡𝑡

𝜕𝜕𝑡𝑡


Recapping, cont.:

−𝛻𝛻2Φ 𝒓𝒓, 𝑡𝑡 −𝜕𝜕𝜕𝜕𝑡𝑡



+ 𝜵𝜵 𝜵𝜵.𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕Φ 𝒓𝒓, 𝑡𝑡

𝜕𝜕𝑡𝑡= 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓, 𝑡𝑡

𝜵𝜵.𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕𝜕𝜕𝑡𝑡Φ 𝒓𝒓, 𝑡𝑡 = 0



−𝛻𝛻2Φ 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕2Φ 𝒓𝒓, 𝑡𝑡


Lorenz gauge:

= 0


In addition to the Lorenz gauge, there are other possible choices to

simplify the equations below !!

−𝛻𝛻2Φ 𝒓𝒓, 𝑡𝑡 −𝜕𝜕𝜕𝜕𝑡𝑡



+ 𝜵𝜵 𝜵𝜵.𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕𝜕𝜕𝑡𝑡Φ 𝒓𝒓, 𝑡𝑡 = 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓, 𝑡𝑡


The Coulomb gauge

−𝛻𝛻2Φ 𝒓𝒓, 𝑡𝑡 −𝜕𝜕𝜕𝜕𝑡𝑡



+ 𝜵𝜵 𝜵𝜵.𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕𝜕𝜕𝑡𝑡Φ 𝒓𝒓, 𝑡𝑡 = 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓, 𝑡𝑡

𝜵𝜵.𝑨𝑨 𝒓𝒓, 𝑡𝑡 = 0

−𝛻𝛻2Φ 𝒓𝒓, 𝑡𝑡 =𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜖𝜖0

−𝛻𝛻𝟐𝟐𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕2𝑨𝑨 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡2 + 𝜇𝜇𝑜𝑜 𝜖𝜖0 𝜵𝜵

𝜕𝜕𝜕𝜕𝑡𝑡 Φ 𝒓𝒓, 𝑡𝑡 = 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓, 𝑡𝑡

or radiation gauge or transverse gauge

= 0

= 0


Φ 𝒓𝒓, 𝑡𝑡 =1

4 𝜋𝜋 𝜖𝜖0�−∞

+∞𝜌𝜌 𝒓𝒓′, 𝑡𝑡𝒓𝒓 − 𝒓𝒓′


−𝛻𝛻2Φ 𝒓𝒓, 𝑡𝑡 =𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜖𝜖0


−𝛻𝛻𝟐𝟐𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕2𝑨𝑨 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡2 = 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓, 𝑡𝑡 − 𝜖𝜖0 𝜵𝜵

𝜕𝜕𝜕𝜕𝑡𝑡 Φ 𝒓𝒓, 𝑡𝑡

= 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓, 𝑡𝑡 −1

4 𝜋𝜋 𝜵𝜵𝜕𝜕𝜕𝜕𝑡𝑡

�−∞



= 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓, 𝑡𝑡 +1

4 𝜋𝜋 𝜵𝜵 �−∞

+∞𝜵𝜵𝑞. 𝑱𝑱 𝒓𝒓𝑞, 𝑡𝑡𝒓𝒓 − 𝒓𝒓′


−𝛻𝛻𝟐𝟐𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕2𝑨𝑨 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡2 = 𝜇𝜇𝑜𝑜 𝑱𝑱𝑻𝑻 𝒓𝒓, 𝑡𝑡

𝜵𝜵. 𝑱𝑱𝑻𝑻 𝒓𝒓, 𝑡𝑡 = 0𝑱𝑱𝑻𝑻 𝒓𝒓, 𝑡𝑡 ≡ 𝑱𝑱 𝒓𝒓, 𝑡𝑡 +1

4 𝜋𝜋 𝜵𝜵 �−∞

+∞𝛻𝛻𝑞. 𝑱𝑱 𝒓𝒓𝑞, 𝑡𝑡𝒓𝒓 − 𝒓𝒓′ 𝑑𝑑𝑑𝑑𝑞


4 𝜋𝜋 𝜖𝜖0�−∞




𝑩𝑩 𝒓𝒓, 𝒕𝒕 = 𝛻𝛻 × 𝑨𝑨 𝒓𝒓, 𝑡𝑡

𝑬𝑬 𝒓𝒓, 𝑡𝑡 = −𝜕𝜕𝑨𝑨 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

−𝛻𝛻𝟐𝟐𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕2𝑨𝑨 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡2 = 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓, 𝑡𝑡 − 𝜖𝜖0 𝜵𝜵

𝜕𝜕𝜕𝜕𝑡𝑡 Φ 𝒓𝒓, 𝑡𝑡

The Coulomb gauge 𝜵𝜵.𝑨𝑨 𝒓𝒓, 𝑡𝑡 = 0is particularly useful in the absence of

charges 𝜌𝜌 𝒓𝒓, 𝑡𝑡 = 0 and currents 𝑱𝑱 𝒓𝒓, 𝑡𝑡 = 0


4 𝜋𝜋 𝜖𝜖0�−∞


𝑑𝑑𝑑𝑑𝑞 = 0= 0

= 0= 0

= 0

−𝛻𝛻𝟐𝟐𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕2𝑨𝑨 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡2 = 0


Conservation Theorems


Energy and Power

𝛁𝛁 × 𝑬𝑬 𝒓𝒓, 𝑡𝑡 = −𝜕𝜕𝑩𝑩 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

𝛁𝛁 × 𝑩𝑩 𝒓𝒓, 𝑡𝑡 = 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕𝑬𝑬 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

𝑩𝑩 𝒓𝒓, 𝑡𝑡 .

𝑬𝑬 𝒓𝒓, 𝑡𝑡 .

𝑩𝑩 𝒓𝒓, 𝑡𝑡 . 𝛁𝛁 × 𝑬𝑬 𝒓𝒓, 𝑡𝑡 − 𝑬𝑬 𝒓𝒓, 𝑡𝑡 . 𝛁𝛁 × 𝑩𝑩 𝒓𝒓, 𝑡𝑡 =

= −𝑩𝑩 𝒓𝒓, 𝑡𝑡 .𝜕𝜕𝑩𝑩 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

−𝜇𝜇𝑜𝑜 𝑬𝑬 𝒓𝒓, 𝑡𝑡 . 𝑱𝑱 𝒓𝒓, 𝑡𝑡 − 𝜇𝜇𝑜𝑜 𝜖𝜖0 𝑬𝑬 𝒓𝒓, 𝑡𝑡 .𝜕𝜕𝑬𝑬 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡


𝛻𝛻. 𝒂𝒂 × 𝒃𝒃 = 𝒃𝒃 .𝛻𝛻 × 𝒂𝒂 − 𝒂𝒂 . 𝛻𝛻 × 𝒃𝒃HW:

= 𝑩𝑩 𝒓𝒓, 𝑡𝑡 . 𝛻𝛻 × 𝑬𝑬 𝒓𝒓, 𝑡𝑡 − 𝑬𝑬 𝒓𝒓, 𝑡𝑡 . 𝛁𝛁 × 𝑩𝑩 𝒓𝒓, 𝑡𝑡

= −12𝜕𝜕 𝑩𝑩 𝒓𝒓, 𝑡𝑡 2

𝜕𝜕𝑡𝑡− 𝜇𝜇𝑜𝑜 𝑬𝑬 𝒓𝒓, 𝑡𝑡 . 𝑱𝑱 𝒓𝒓, 𝑡𝑡 − 𝜇𝜇𝑜𝑜 𝜖𝜖0

12𝜕𝜕 𝑬𝑬 𝒓𝒓, 𝑡𝑡 2

𝜕𝜕𝑡𝑡

= −𝑩𝑩 𝒓𝒓, 𝑡𝑡 .𝜕𝜕𝑩𝑩 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

− 𝜇𝜇𝑜𝑜 𝑬𝑬 𝒓𝒓, 𝑡𝑡 . 𝑱𝑱 𝒓𝒓, 𝑡𝑡 − 𝜇𝜇𝑜𝑜 𝜖𝜖0 𝑬𝑬 𝒓𝒓, 𝑡𝑡 .𝜕𝜕𝑬𝑬 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

𝛻𝛻. 𝑬𝑬 𝒓𝒓, 𝑡𝑡 ×𝑩𝑩 𝒓𝒓, 𝑡𝑡𝜇𝜇𝑜𝑜

+𝜖𝜖02𝜕𝜕 𝑬𝑬 𝒓𝒓, 𝑡𝑡 2

𝜕𝜕𝑡𝑡+

12 𝜇𝜇𝑜𝑜

𝜕𝜕 𝑩𝑩 𝒓𝒓, 𝑡𝑡 2

𝜕𝜕𝑡𝑡+ 𝑬𝑬 𝒓𝒓, 𝑡𝑡 . 𝑱𝑱 𝒓𝒓, 𝑡𝑡 = 0

𝛻𝛻. 𝑬𝑬 𝒓𝒓, 𝑡𝑡 × 𝑩𝑩 𝒓𝒓, 𝑡𝑡


𝛻𝛻. 𝑬𝑬 𝒓𝒓, 𝑡𝑡 ×𝑩𝑩 𝒓𝒓, 𝑡𝑡𝜇𝜇𝑜𝑜

+𝜖𝜖02𝜕𝜕 𝑬𝑬 𝒓𝒓, 𝑡𝑡 2

𝜕𝜕𝑡𝑡+

12 𝜇𝜇𝑜𝑜

𝜕𝜕 𝑩𝑩 𝒓𝒓, 𝑡𝑡 2

𝜕𝜕𝑡𝑡+ 𝑬𝑬 𝒓𝒓, 𝑡𝑡 . 𝑱𝑱 𝒓𝒓, 𝑡𝑡 = 0

𝛻𝛻.𝓢𝓢 +𝜕𝜕𝐹𝐹𝐸𝐸𝜕𝜕𝑡𝑡

+𝜕𝜕𝐹𝐹𝐵𝐵𝜕𝜕𝑡𝑡

+𝜕𝜕𝑤𝑤𝜕𝜕𝑡𝑡

= 0

𝓢𝓢 ≡ 𝑬𝑬 𝒓𝒓, 𝑡𝑡 ×𝑩𝑩 𝒓𝒓, 𝑡𝑡𝜇𝜇𝑜𝑜

𝐹𝐹𝐸𝐸 =𝜖𝜖02 𝑬𝑬 𝒓𝒓, 𝑡𝑡 2

𝐹𝐹𝐵𝐵 =1

2 𝜇𝜇𝑜𝑜𝑩𝑩 𝒓𝒓, 𝑡𝑡 2

𝜕𝜕𝑤𝑤𝜕𝜕𝑡𝑡 = 𝑬𝑬 𝒓𝒓, 𝑡𝑡 . 𝑱𝑱 𝒓𝒓, 𝑡𝑡

Differential form of conservation of energy

(per-unit-time)

Poynting vector


�𝑉𝑉

𝛻𝛻.𝓢𝓢 +𝜕𝜕𝐹𝐹𝐸𝐸𝜕𝜕𝑡𝑡

+𝜕𝜕𝐹𝐹𝐵𝐵𝜕𝜕𝑡𝑡

+𝜕𝜕𝑤𝑤𝜕𝜕𝑡𝑡

𝑑𝑑𝑑𝑑 = 0

�𝑆𝑆𝓢𝓢 .𝒅𝒅𝒅𝒅 +

𝑑𝑑𝑈𝑈𝐸𝐸𝑑𝑑𝑡𝑡

+𝑑𝑑𝑈𝑈𝐵𝐵𝑑𝑑𝑡𝑡

+𝑑𝑑𝑊𝑊𝑑𝑑𝑡𝑡

= 0


𝑑𝑑𝑑𝑑𝑡𝑡�

𝑉𝑉𝐹𝐹𝐸𝐸 + 𝐹𝐹𝐵𝐵 + 𝑤𝑤 𝑑𝑑𝑑𝑑 = 0

Integral form of conservation of

energy (per-unit-time)


𝛁𝛁.𝓢𝓢 +𝜕𝜕𝜕𝜕𝑡𝑡

𝐹𝐹𝐸𝐸 + 𝐹𝐹𝐵𝐵 + 𝑤𝑤 = 0

𝓢𝓢 ≡ 𝑬𝑬 𝒓𝒓, 𝑡𝑡 ×𝑩𝑩 𝒓𝒓, 𝑡𝑡𝜇𝜇𝑜𝑜

𝐹𝐹𝐸𝐸 =𝜖𝜖02𝑬𝑬 𝒓𝒓, 𝑡𝑡 2 𝐹𝐹𝐵𝐵 =

12 𝜇𝜇𝑜𝑜

𝑩𝑩 𝒓𝒓, 𝑡𝑡 2

𝑤𝑤 = 𝑬𝑬 𝒓𝒓, 𝑡𝑡 . 𝑱𝑱 𝒓𝒓, 𝑡𝑡

Recapping: Power Conservation


𝑑𝑑𝑈𝑈𝐸𝐸𝑑𝑑𝑡𝑡

+𝑑𝑑𝑈𝑈𝐵𝐵𝑑𝑑𝑡𝑡

+𝑑𝑑𝑊𝑊𝑑𝑑𝑡𝑡

= 0


HW: From Maxwell’s equations, prove the following relation:

𝜖𝜖0 𝑬𝑬 𝒓𝒓, 𝑡𝑡 × 𝛁𝛁 × 𝑬𝑬 𝒓𝒓, 𝑡𝑡 − 𝑬𝑬 𝒓𝒓, 𝑡𝑡 𝛁𝛁.𝑬𝑬 𝒓𝒓, 𝑡𝑡

+1𝜇𝜇𝑜𝑜

𝑩𝑩 𝒓𝒓, 𝑡𝑡 × 𝛁𝛁 × 𝑩𝑩 𝒓𝒓, 𝑡𝑡 − 𝑩𝑩 𝒓𝒓, 𝑡𝑡 𝛁𝛁.𝑩𝑩 𝒓𝒓, 𝑡𝑡

+ 𝜌𝜌 𝒓𝒓, 𝑡𝑡 𝑬𝑬 𝒓𝒓, 𝑡𝑡 + 𝑱𝑱 𝒓𝒓, 𝑡𝑡 × 𝑩𝑩 𝒓𝒓, 𝑡𝑡

+𝜕𝜕𝜕𝜕𝑡𝑡

𝜖𝜖0 𝜇𝜇𝑜𝑜 𝑬𝑬 𝒓𝒓, 𝑡𝑡 ×𝑩𝑩 𝒓𝒓, 𝑡𝑡𝜇𝜇𝑜𝑜

= 0


HW: Show that

= �𝑗𝑗=1

3𝜕𝜕𝜕𝜕𝐹𝐹𝑗𝑗

𝜖𝜖012𝑬𝑬 2𝛿𝛿𝑖𝑖𝑗𝑗 − 𝐸𝐸𝑖𝑖 𝐸𝐸𝑗𝑗 +

1𝜇𝜇𝑜𝑜

12𝑩𝑩 2𝛿𝛿𝑖𝑖𝑗𝑗 − 𝐵𝐵𝑖𝑖 𝐵𝐵𝑗𝑗

𝜖𝜖0 𝑬𝑬 𝒓𝒓, 𝑡𝑡 × 𝛁𝛁 × 𝑬𝑬 𝒓𝒓, 𝑡𝑡 − 𝑬𝑬 𝒓𝒓, 𝑡𝑡 𝛁𝛁.𝑬𝑬 𝒓𝒓, 𝑡𝑡 𝑖𝑖 +

+1𝜇𝜇𝑜𝑜

𝑩𝑩 𝒓𝒓, 𝑡𝑡 × 𝛁𝛁 × 𝑩𝑩 𝒓𝒓, 𝑡𝑡 − 𝑩𝑩 𝒓𝒓, 𝑡𝑡 𝛁𝛁.𝑩𝑩 𝒓𝒓, 𝑡𝑡 𝑖𝑖 =


Linear Momentum

𝛁𝛁 × 𝑬𝑬 𝒓𝒓, 𝑡𝑡 +𝜕𝜕𝑩𝑩 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

= 0

𝛁𝛁 × 𝑩𝑩 𝒓𝒓, 𝑡𝑡 − 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓, 𝑡𝑡 − 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕𝑬𝑬 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

= 01𝜇𝜇𝑜𝑜𝑩𝑩 𝒓𝒓, 𝑡𝑡 ×

− 𝜖𝜖0 𝑬𝑬 𝒓𝒓, 𝑡𝑡 𝛁𝛁.𝑬𝑬 𝒓𝒓, 𝑡𝑡 −𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜖𝜖0

= 0

𝛁𝛁.𝑩𝑩 𝒓𝒓, 𝑡𝑡 = 0−1𝜇𝜇𝑜𝑜𝑩𝑩 𝒓𝒓, 𝑡𝑡

𝜖𝜖0 𝑬𝑬 𝒓𝒓, 𝑡𝑡 ×



+1𝜇𝜇𝑜𝑜


+ 𝜌𝜌 𝒓𝒓, 𝑡𝑡 𝑬𝑬 𝒓𝒓, 𝑡𝑡 + 𝑱𝑱 𝒓𝒓, 𝑡𝑡 × 𝑩𝑩 𝒓𝒓, 𝑡𝑡

+𝜕𝜕𝜕𝜕𝑡𝑡


= 0


𝜌𝜌 𝒓𝒓, 𝑡𝑡 𝑬𝑬 𝒓𝒓, 𝑡𝑡 + 𝑱𝑱 𝒓𝒓, 𝑡𝑡 × 𝑩𝑩 𝒓𝒓, 𝑡𝑡𝑑𝑑𝑑𝑑 = 𝑑𝑑𝑭𝑭𝐿𝐿 𝒓𝒓, 𝑡𝑡

𝜌𝜌 𝒓𝒓, 𝑡𝑡 𝑬𝑬 𝒓𝒓, 𝑡𝑡 + 𝑱𝑱 𝒓𝒓, 𝑡𝑡 × 𝑩𝑩 𝒓𝒓, 𝑡𝑡 =𝑑𝑑𝑭𝑭𝐿𝐿 𝒓𝒓, 𝑡𝑡

𝑑𝑑𝑑𝑑the Lorentz force (per-unit-volume) due to the fields acting on charged particles

(charges and currents)the rate of

change of the linear momentum (per-unit-volume) of the charged

particles

=𝜕𝜕𝜕𝜕𝑡𝑡𝒑𝒑 𝒓𝒓, 𝑡𝑡≡ 𝒇𝒇𝐿𝐿 𝒓𝒓, 𝑡𝑡


+𝜕𝜕𝜕𝜕𝑡𝑡𝒑𝒑 𝒓𝒓, 𝑡𝑡

+𝜕𝜕𝜕𝜕𝑡𝑡


= 0


+1𝜇𝜇𝑜𝑜


+ 𝜌𝜌 𝒓𝒓, 𝑡𝑡 𝑬𝑬 𝒓𝒓, 𝑡𝑡 + 𝑱𝑱 𝒓𝒓, 𝑡𝑡 × 𝑩𝑩 𝒓𝒓, 𝑡𝑡+ 𝒇𝒇𝐿𝐿 𝒓𝒓, 𝑡𝑡


𝒈𝒈 𝒓𝒓, 𝑡𝑡 ≡ 𝜖𝜖0 𝜇𝜇𝑜𝑜 𝑬𝑬 𝒓𝒓, 𝑡𝑡 ×𝑩𝑩 𝒓𝒓, 𝑡𝑡𝜇𝜇𝑜𝑜

Linear momentum (per-unit-volume) associated withthe E & B fields

𝜕𝜕𝜕𝜕𝑡𝑡


=𝜕𝜕𝜕𝜕𝑡𝑡𝒈𝒈 𝒓𝒓, 𝑡𝑡

= 𝜖𝜖0 𝜇𝜇𝑜𝑜 𝓢𝓢


+𝜕𝜕𝒑𝒑 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

+𝜕𝜕𝒈𝒈 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

= 0


+1𝜇𝜇𝑜𝑜


+𝜕𝜕𝜕𝜕𝑡𝑡



+𝜕𝜕𝑝𝑝𝑖𝑖 𝒓𝒓, 𝑡𝑡

𝜕𝜕𝑡𝑡

+𝜕𝜕𝑔𝑔𝑖𝑖 𝒓𝒓, 𝑡𝑡

𝜕𝜕𝑡𝑡

= 0

𝜖𝜖0 𝑬𝑬 𝒓𝒓, 𝑡𝑡 × 𝛁𝛁 × 𝑬𝑬 𝒓𝒓, 𝑡𝑡 − 𝑬𝑬 𝒓𝒓, 𝑡𝑡 𝛁𝛁.𝑬𝑬 𝒓𝒓, 𝑡𝑡 𝑖𝑖

+1𝜇𝜇𝑜𝑜

𝑩𝑩 𝒓𝒓, 𝑡𝑡 × 𝛁𝛁 × 𝑩𝑩 𝒓𝒓, 𝑡𝑡 − 𝑩𝑩 𝒓𝒓, 𝑡𝑡 𝛁𝛁.𝑩𝑩 𝒓𝒓, 𝑡𝑡 𝑖𝑖

Let’s calculate one Cartesian component:


𝑬𝑬 × 𝛁𝛁 × 𝑬𝑬 − 𝑬𝑬 𝛁𝛁.𝑬𝑬 𝑖𝑖

= 𝜖𝜖𝑖𝑖𝑗𝑗𝑖𝑖 𝐸𝐸𝑗𝑗 𝜖𝜖𝑖𝑖𝑎𝑎𝑘𝑘𝜕𝜕𝐸𝐸𝑘𝑘𝜕𝜕𝐹𝐹𝑎𝑎

− 𝐸𝐸𝑖𝑖𝜕𝜕𝐸𝐸𝑗𝑗𝜕𝜕𝐹𝐹𝑗𝑗

𝑄𝑄𝑖𝑖 = 𝛁𝛁 × 𝑬𝑬 𝑖𝑖

𝑬𝑬 × 𝑸𝑸 𝑖𝑖

𝑬𝑬 𝛁𝛁.𝑬𝑬 𝑖𝑖

= 𝜖𝜖𝑖𝑖𝑗𝑗𝑖𝑖 𝐸𝐸𝑗𝑗 𝛁𝛁 × 𝑬𝑬 𝑖𝑖 − 𝐸𝐸𝑖𝑖𝜕𝜕𝐸𝐸𝑗𝑗𝜕𝜕𝐹𝐹𝑗𝑗

= 𝜖𝜖𝑖𝑖𝑗𝑗𝑖𝑖 𝐸𝐸𝑗𝑗 𝑄𝑄𝑖𝑖

= 𝜖𝜖𝑖𝑖𝑎𝑎𝑘𝑘𝜕𝜕𝐸𝐸𝑘𝑘𝜕𝜕𝐹𝐹𝑎𝑎

= 𝐸𝐸𝑖𝑖𝜕𝜕𝐸𝐸𝑗𝑗𝜕𝜕𝐹𝐹𝑗𝑗

𝜖𝜖𝑖𝑖𝑗𝑗𝑖𝑖

otherwise

(1,2,3) or (2,3,1) or (3,1,2)

(1,3,2) or (3,2,1) or (2,1,3)

= 0

= + 1

= - 1

= �𝑗𝑗=1

3

�𝑖𝑖=1

3

𝜖𝜖𝑖𝑖𝑗𝑗𝑖𝑖 𝐸𝐸𝑗𝑗 𝑄𝑄𝑖𝑖

= �𝑘𝑘=1

3

�𝑎𝑎=1

3

𝜖𝜖𝑖𝑖𝑎𝑎𝑘𝑘𝜕𝜕𝐸𝐸𝑘𝑘𝜕𝜕𝐹𝐹𝑎𝑎

= 𝐸𝐸𝑖𝑖�𝑗𝑗=1

3𝜕𝜕𝐸𝐸𝑗𝑗𝜕𝜕𝐹𝐹𝑗𝑗

𝑬𝑬 𝒓𝒓, 𝑡𝑡 × 𝛁𝛁 × 𝑬𝑬 𝒓𝒓, 𝑡𝑡 − 𝑬𝑬 𝒓𝒓, 𝑡𝑡 𝛁𝛁.𝑬𝑬 𝒓𝒓, 𝑡𝑡 𝑖𝑖


𝜖𝜖𝑖𝑖𝑗𝑗𝑖𝑖 𝜖𝜖𝑖𝑖𝑎𝑎𝑘𝑘

= 𝛿𝛿𝑖𝑖𝑎𝑎 𝛿𝛿𝑗𝑗𝑘𝑘

𝑬𝑬 × 𝛁𝛁 × 𝑬𝑬 − 𝑬𝑬 𝛁𝛁.𝑬𝑬 𝑖𝑖 = 𝜖𝜖𝑖𝑖𝑗𝑗𝑖𝑖 𝜖𝜖𝑖𝑖𝑎𝑎𝑘𝑘 𝐸𝐸𝑗𝑗𝜕𝜕𝐸𝐸𝑘𝑘𝜕𝜕𝐹𝐹𝑎𝑎


= 𝐸𝐸𝑗𝑗𝜕𝜕𝐸𝐸𝑗𝑗𝜕𝜕𝐹𝐹𝑖𝑖

−𝜕𝜕𝐸𝐸𝑖𝑖𝜕𝜕𝐹𝐹𝑗𝑗


= 𝛿𝛿𝑖𝑖𝑎𝑎 𝛿𝛿𝑗𝑗𝑘𝑘 − 𝛿𝛿𝑖𝑖𝑘𝑘 𝛿𝛿𝑗𝑗𝑎𝑎 𝐸𝐸𝑗𝑗𝜕𝜕𝐸𝐸𝑘𝑘𝜕𝜕𝐹𝐹𝑎𝑎


− 𝛿𝛿𝑖𝑖𝑘𝑘 𝛿𝛿𝑗𝑗𝑎𝑎

= 𝜖𝜖𝑖𝑖𝑖𝑖𝑗𝑗 𝜖𝜖𝑖𝑖𝑎𝑎𝑘𝑘


𝑬𝑬 × 𝛁𝛁 × 𝑬𝑬 − 𝑬𝑬 𝛁𝛁.𝑬𝑬 𝑖𝑖 = 𝐸𝐸𝑗𝑗𝜕𝜕𝐸𝐸𝑗𝑗𝜕𝜕𝐹𝐹𝑖𝑖

−𝜕𝜕𝐸𝐸𝑖𝑖𝜕𝜕𝐹𝐹𝑗𝑗


= 𝐸𝐸𝑗𝑗𝜕𝜕𝐸𝐸𝑗𝑗𝜕𝜕𝐹𝐹𝑖𝑖

− 𝐸𝐸𝑗𝑗𝜕𝜕𝐸𝐸𝑖𝑖𝜕𝜕𝐹𝐹𝑗𝑗


=12𝜕𝜕 𝐸𝐸𝑗𝑗 𝐸𝐸𝑗𝑗𝜕𝜕𝐹𝐹𝑖𝑖

−𝜕𝜕 𝐸𝐸𝑖𝑖 𝐸𝐸𝑗𝑗𝜕𝜕𝐹𝐹𝑗𝑗

=12𝜕𝜕 𝑬𝑬 2

𝜕𝜕𝐹𝐹𝑗𝑗𝛿𝛿𝑖𝑖𝑗𝑗 −

𝜕𝜕 𝐸𝐸𝑖𝑖 𝐸𝐸𝑗𝑗𝜕𝜕𝐹𝐹𝑗𝑗

=𝜕𝜕𝜕𝜕𝐹𝐹𝑗𝑗

12𝑬𝑬 2𝛿𝛿𝑖𝑖𝑗𝑗 − 𝐸𝐸𝑖𝑖 𝐸𝐸𝑗𝑗


+𝜕𝜕𝑝𝑝𝑖𝑖 𝒓𝒓, 𝑡𝑡

𝜕𝜕𝑡𝑡

+𝜕𝜕𝑔𝑔𝑖𝑖 𝒓𝒓, 𝑡𝑡

𝜕𝜕𝑡𝑡

= 0

𝜖𝜖0 𝑬𝑬 𝒓𝒓, 𝑡𝑡 × 𝛁𝛁 × 𝑬𝑬 𝒓𝒓, 𝑡𝑡 − 𝑬𝑬 𝒓𝒓, 𝑡𝑡 𝛁𝛁.𝑬𝑬 𝒓𝒓, 𝑡𝑡 𝑖𝑖

+1𝜇𝜇𝑜𝑜

𝑩𝑩 𝒓𝒓, 𝑡𝑡 × 𝛁𝛁 × 𝑩𝑩 𝒓𝒓, 𝑡𝑡 − 𝑩𝑩 𝒓𝒓, 𝑡𝑡 𝛁𝛁.𝑩𝑩 𝒓𝒓, 𝑡𝑡 𝑖𝑖


+𝜕𝜕𝑝𝑝𝑖𝑖𝜕𝜕𝑡𝑡

+𝜕𝜕𝑔𝑔𝑖𝑖𝜕𝜕𝑡𝑡

= 0

�𝑗𝑗=1


𝜖𝜖012𝑬𝑬 2𝛿𝛿𝑖𝑖𝑗𝑗 − 𝐸𝐸𝑖𝑖 𝐸𝐸𝑗𝑗

+�𝑗𝑗=1


1𝜇𝜇𝑜𝑜

12𝑩𝑩 2𝛿𝛿𝑖𝑖𝑗𝑗 − 𝐵𝐵𝑖𝑖 𝐵𝐵𝑗𝑗

�𝑗𝑗=1

3𝜕𝜕𝑇𝑇𝑖𝑖𝑗𝑗𝜕𝜕𝐹𝐹𝑗𝑗

𝑇𝑇𝑖𝑖𝑗𝑗 ≡ 𝜖𝜖012𝑬𝑬 2𝛿𝛿𝑖𝑖𝑗𝑗 − 𝐸𝐸𝑖𝑖 𝐸𝐸𝑗𝑗 +

1𝜇𝜇𝑜𝑜

12𝑩𝑩 2 𝛿𝛿𝑖𝑖𝑗𝑗 − 𝐵𝐵𝑖𝑖 𝐵𝐵𝑗𝑗

Maxwell’s stress tensor


�𝑗𝑗=1




= 0

�𝑉𝑉�𝑗𝑗=1


𝑑𝑑𝑑𝑑 + �𝑉𝑉

𝜕𝜕𝑔𝑔𝑖𝑖𝜕𝜕𝑡𝑡



�𝑆𝑆�𝑗𝑗=1

3

𝑇𝑇𝑖𝑖𝑗𝑗 𝑑𝑑𝑆𝑆𝑗𝑗 +𝑑𝑑𝑑𝑑𝑡𝑡�

𝑉𝑉𝑔𝑔𝑖𝑖 + 𝑝𝑝𝑖𝑖 𝑑𝑑𝑑𝑑 = 0

−𝐹𝐹𝑖𝑖


𝐹𝐹𝑖𝑖 = −�𝑆𝑆�𝑗𝑗=1

3

𝑇𝑇𝑖𝑖𝑗𝑗 𝑑𝑑𝑆𝑆𝑗𝑗

𝑭𝑭 = −�𝑆𝑆𝑻𝑻 . �𝒏𝒏 𝑑𝑑𝑆𝑆

𝑑𝑑𝑭𝑭𝑑𝑑𝑆𝑆

= − 𝑻𝑻 . �𝒏𝒏

= −�𝑆𝑆�𝑗𝑗=1

3

𝑇𝑇𝑖𝑖𝑗𝑗 �𝑛𝑛𝑗𝑗 𝑑𝑑𝑆𝑆�𝒏𝒏

unit vector normal to

the surface

𝑑𝑑𝐹𝐹𝑖𝑖𝑑𝑑𝑆𝑆

= −�𝑗𝑗=1

3

𝑇𝑇𝑖𝑖𝑗𝑗 �𝑛𝑛𝑗𝑗



1𝜇𝜇𝑜𝑜


= 𝜖𝜖0 −12𝑬𝑬 2 �𝒏𝒏 + 𝑬𝑬 𝑬𝑬. �𝒏𝒏 +

1𝜇𝜇𝑜𝑜

−12𝑩𝑩 2 �𝒏𝒏 + 𝑩𝑩 𝑩𝑩. �𝒏𝒏


= − 𝑻𝑻 . �𝒏𝒏


= −𝜖𝜖02𝑬𝑬 2 +

12 𝜇𝜇𝑜𝑜

𝑩𝑩 2 �𝒏𝒏 + 𝜖𝜖0 𝑬𝑬 𝑬𝑬. �𝒏𝒏 +1𝜇𝜇𝑜𝑜𝑩𝑩 𝑩𝑩. �𝒏𝒏

along the surface normal and always compressingthe surface inwards

along the fieldsand always tensioningthe surface outwards


Recapping:

+𝜕𝜕𝒑𝒑𝜕𝜕𝑡𝑡

+𝜕𝜕𝒈𝒈𝜕𝜕𝑡𝑡

= 0

𝜖𝜖0 𝑬𝑬 × 𝛁𝛁 × 𝑬𝑬 − 𝑬𝑬 𝛁𝛁.𝑬𝑬

+1𝜇𝜇𝑜𝑜

𝑩𝑩 × 𝛁𝛁 × 𝑩𝑩 −𝑩𝑩 𝛁𝛁.𝑩𝑩


Recapping, cont.:

𝜕𝜕𝒑𝒑𝜕𝜕𝑡𝑡

= 𝜌𝜌 𝑬𝑬 + 𝑱𝑱 × 𝑩𝑩 =𝑑𝑑𝑭𝑭𝐿𝐿𝑑𝑑𝑑𝑑

≡ 𝒇𝒇𝐿𝐿

𝜕𝜕𝒈𝒈𝜕𝜕𝑡𝑡

≡ 𝜖𝜖0 𝜇𝜇𝑜𝑜𝜕𝜕𝜕𝜕𝑡𝑡

𝑬𝑬 ×𝑩𝑩𝜇𝜇𝑜𝑜

= 𝜖𝜖0 𝜇𝜇𝑜𝑜𝜕𝜕𝓢𝓢𝜕𝜕𝑡𝑡

𝜖𝜖0 𝑬𝑬 × 𝛁𝛁 × 𝑬𝑬 − 𝑬𝑬 𝛁𝛁.𝑬𝑬 𝑖𝑖 +1𝜇𝜇𝑜𝑜

𝑩𝑩 × 𝛁𝛁 × 𝑩𝑩 −𝑩𝑩 𝛁𝛁.𝑩𝑩 𝑖𝑖 ≡�𝑗𝑗=1



1𝜇𝜇𝑜𝑜



Recapping, cont.:

�𝑉𝑉

𝛁𝛁.𝑻𝑻 +𝜕𝜕𝒑𝒑𝜕𝜕𝑡𝑡



�𝑆𝑆𝑻𝑻.𝒅𝒅𝒅𝒅 +

𝑑𝑑𝑑𝑑𝑡𝑡

𝑷𝑷 + 𝑮𝑮 = 0

−𝑻𝑻: the force (per-unit-area) creates a change in the total linear momentum

from both fields and particles

𝛁𝛁.𝑻𝑻 +𝜕𝜕𝒑𝒑𝜕𝜕𝑡𝑡


= 0

Conservation of Linear Momentum

�𝑗𝑗=1




= 0


Recapping, cont.:

− 𝑇𝑇𝑖𝑖𝑗𝑗 ≡ −𝜖𝜖02𝑬𝑬 2 +

12 𝜇𝜇𝑜𝑜

𝑩𝑩 2 𝛿𝛿𝑖𝑖𝑗𝑗 + 𝜖𝜖0 𝐸𝐸𝑖𝑖 𝐸𝐸𝑗𝑗 +1𝜇𝜇𝑜𝑜𝐵𝐵𝑖𝑖 𝐵𝐵𝑗𝑗

𝑑𝑑𝐹𝐹𝑖𝑖 = −�𝑗𝑗=1

3

𝑇𝑇𝑖𝑖𝑗𝑗 𝑑𝑑𝑆𝑆𝑗𝑗


= −𝜖𝜖02𝑬𝑬 2 +

12 𝜇𝜇𝑜𝑜

𝑩𝑩 2 �𝒏𝒏 + 𝜖𝜖0 𝑬𝑬 𝑬𝑬. �𝒏𝒏 +1𝜇𝜇𝑜𝑜𝑩𝑩 𝑩𝑩. �𝒏𝒏

along the surface normal and always compressingthe surface inwards

along the fieldsand always tensioningthe surface outwards


𝑬𝑬 ⊥ �𝒏𝒏


= −𝜖𝜖02𝑬𝑬 2 �𝒏𝒏

𝑩𝑩 ⊥ �𝒏𝒏


= −1

2 𝜇𝜇𝑜𝑜𝑩𝑩 2�𝒏𝒏

𝑩𝑩 = 𝟎𝟎 𝑬𝑬 = 0

𝑑𝑑𝑭𝑭�𝒏𝒏 𝑑𝑑𝑭𝑭

�𝒏𝒏


𝑑𝑑𝑭𝑭�𝒏𝒏

�𝒏𝒏𝑑𝑑𝑭𝑭

crashing can

https://www.youtube.com/watch?v=YFTH8ywXaZs

https://www.youtube.com/watch?v=YFTH8ywXaZs



𝑬𝑬 ∥ �𝒏𝒏 or −�𝒏𝒏𝑩𝑩 = 𝟎𝟎


= +𝜖𝜖02𝑬𝑬 2 �𝒏𝒏


= +1

2 𝜇𝜇𝑜𝑜𝑩𝑩 2�𝒏𝒏

𝑩𝑩 ∥ �𝒏𝒏 or − �𝒏𝒏𝑬𝑬 = 𝟎𝟎






𝑑𝑑𝑭𝑭

�𝒏𝒏

�𝒏𝒏

𝑑𝑑𝑭𝑭

𝑑𝑑𝑭𝑭

�𝒏𝒏

�𝒏𝒏

𝑑𝑑𝑭𝑭


Angular Momentum

𝜵𝜵.𝑻𝑻 +𝜕𝜕𝒈𝒈𝜕𝜕𝑡𝑡

+ 𝒇𝒇𝐿𝐿 = 0𝒓𝒓 ×

𝒓𝒓 × 𝜵𝜵.𝑻𝑻 + 𝒓𝒓 ×𝜕𝜕𝒈𝒈𝜕𝜕𝑡𝑡

+ = 0𝜕𝜕𝒄𝒄𝑘𝑘𝑖𝑖𝑚𝑚𝑚

𝜕𝜕𝑡𝑡

𝒏𝒏𝑘𝑘𝑖𝑖𝑚𝑚𝑚 ≡ 𝒓𝒓 × 𝒇𝒇𝐿𝐿 =𝜕𝜕𝒄𝒄𝑘𝑘𝑖𝑖𝑚𝑚𝑚

𝜕𝜕𝑡𝑡

We have found that:

the Lorentz torque (per-unit-volume)due to the fields acting on charged particles (charges and currents)

𝒓𝒓 × 𝒇𝒇𝐿𝐿

the rate of change of the angular momentum (per-unit-volume) of the charged particles


𝜵𝜵. 𝒓𝒓 × 𝑻𝑻 +𝜕𝜕𝒄𝒄𝑖𝑖𝑘𝑘𝜕𝜕𝑡𝑡

+𝜕𝜕𝒄𝒄𝑘𝑘𝑖𝑖𝑚𝑚𝑚

𝜕𝜕𝑡𝑡= 0

𝒓𝒓 × 𝜵𝜵.𝑻𝑻 + 𝒓𝒓 ×𝜕𝜕𝒈𝒈𝜕𝜕𝑡𝑡


𝜕𝜕𝑡𝑡= 0

𝒄𝒄𝑖𝑖𝑘𝑘 ≡ 𝒓𝒓 × 𝒈𝒈

𝒓𝒓 × 𝜵𝜵.𝑻𝑻 +𝜕𝜕𝒄𝒄𝑖𝑖𝑘𝑘𝜕𝜕𝑡𝑡


𝜕𝜕𝑡𝑡= 0

𝒓𝒓 × 𝜵𝜵.𝑻𝑻 = 𝜵𝜵. 𝒓𝒓 × 𝑻𝑻

𝒓𝒓 × 𝜵𝜵.𝑻𝑻 +𝜕𝜕 𝒓𝒓 × 𝒈𝒈

𝜕𝜕𝑡𝑡+𝜕𝜕𝒄𝒄𝑘𝑘𝑖𝑖𝑚𝑚𝑚

𝜕𝜕𝑡𝑡= 0

HW:




𝜕𝜕𝑡𝑡= 0

�𝑉𝑉



𝜕𝜕𝑡𝑡𝑑𝑑𝑑𝑑 = 0

�𝑆𝑆

𝒓𝒓 × 𝑻𝑻 .𝒅𝒅𝒅𝒅 +𝑑𝑑𝑑𝑑𝑡𝑡

𝑳𝑳𝑖𝑖𝑘𝑘 + 𝑳𝑳𝑘𝑘𝑖𝑖𝑚𝑚𝑚 = 0

𝑵𝑵 ≡ 𝒓𝒓 × 𝑻𝑻−𝑵𝑵: the torque (per-unit-area) creates a change in the total angular momentum

from both fields and particles


𝛻𝛻.𝑬𝑬 𝒓𝒓, 𝑡𝑡 =

𝛻𝛻.𝑩𝑩 𝒓𝒓, 𝑡𝑡 = 0

𝛻𝛻 × 𝑬𝑬 𝒓𝒓, 𝑡𝑡 = −𝜕𝜕𝑩𝑩 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

𝛁𝛁 × 𝑩𝑩 𝒓𝒓, 𝑡𝑡 = 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕𝑬𝑬 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

Electromagnetic Waves𝜌𝜌 𝒓𝒓, 𝑡𝑡 = 0 𝑱𝑱 𝒓𝒓, 𝑡𝑡 = 0

+ 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓, 𝑡𝑡

𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜖𝜖0

Consider: &

0



𝜵𝜵 ×

𝛻𝛻 × 𝛻𝛻 × 𝒂𝒂 = 𝛻𝛻 𝛻𝛻.𝒂𝒂 − 𝛻𝛻𝟐𝟐𝒂𝒂HW:

𝜵𝜵 × 𝛁𝛁 × 𝑬𝑬 𝒓𝒓, 𝑡𝑡 = 𝛁𝛁 𝛁𝛁.𝑬𝑬 𝒓𝒓, 𝑡𝑡 − 𝛁𝛁𝟐𝟐𝑬𝑬 𝒓𝒓, 𝑡𝑡


𝛻𝛻.𝑬𝑬 𝒓𝒓, 𝑡𝑡 = 𝟎𝟎

𝛻𝛻𝟐𝟐𝑬𝑬 𝒓𝒓, 𝑡𝑡 = 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕2𝑬𝑬 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡2

= −𝛁𝛁 ×𝜕𝜕𝑩𝑩 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

= −𝜕𝜕𝜕𝜕𝑡𝑡

𝛁𝛁 × 𝑩𝑩 𝒓𝒓, 𝑡𝑡



𝜵𝜵 ×

𝛻𝛻 × 𝛻𝛻 × 𝒂𝒂 = 𝛻𝛻 𝛻𝛻.𝒂𝒂 − 𝛻𝛻𝟐𝟐𝒂𝒂HW:

𝜵𝜵 × 𝛁𝛁 × 𝑩𝑩 𝒓𝒓, 𝑡𝑡 = 𝛁𝛁 𝛁𝛁.𝑩𝑩 𝒓𝒓, 𝑡𝑡 − 𝛁𝛁𝟐𝟐𝑩𝑩 𝒓𝒓, 𝑡𝑡


𝛻𝛻.𝑩𝑩 𝒓𝒓, 𝑡𝑡 = 𝟎𝟎

𝛻𝛻𝟐𝟐𝑩𝑩 𝒓𝒓, 𝑡𝑡 = 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕2𝑩𝑩 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡2

= 𝜇𝜇𝑜𝑜 𝜖𝜖0 𝛁𝛁 ×𝜕𝜕𝑬𝑬 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

= 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕𝜕𝜕𝑡𝑡

𝛁𝛁 × 𝑬𝑬 𝒓𝒓, 𝑡𝑡


𝛻𝛻𝟐𝟐𝑬𝑬 𝒓𝒓, 𝑡𝑡 = 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕2𝑬𝑬 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡2

𝛻𝛻𝟐𝟐𝑩𝑩 𝒓𝒓, 𝑡𝑡 = 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕2𝑩𝑩 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡2

𝑣𝑣 =1𝜇𝜇𝑜𝑜 𝜖𝜖0

≅ 2.99792458 × 108 m/s

Date post:	08-Mar-2021
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Foundations of Electromagnetic Theory 611 spring 19...Electrostatic Theory: 3 charges are not...

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